Progression and Sequence Questions in English

(D) Let the corresponding $A.P.$ have the first term $a$ and common difference $d$.
Since the $n^{th}$ term of $H.P.$ is the reciprocal of the $n^{th}$ term of $A.P.$,we have:
$7^{th}$ term of $A.P. = a + 6d = 10$
$12^{th}$ term of $A.P. = a + 11d = 25$
Subtracting the first equation from the second:
$(a + 11d) - (a + 6d) = 25 - 10$
$5d = 15 \Rightarrow d = 3$
Substituting $d = 3$ into $a + 6d = 10$:
$a + 6(3) = 10 \Rightarrow a + 18 = 10 \Rightarrow a = -8$
The $20^{th}$ term of the $A.P.$ is $T_{20} = a + 19d = -8 + 19(3) = -8 + 57 = 49$.
Therefore,the $20^{th}$ term of the $H.P.$ is the reciprocal of $49$,which is $\frac{1}{49}$.

(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d.$
Since the $n^{th}$ term of a $H.P.$ is the reciprocal of the $n^{th}$ term of an $A.P.,$ we have:
$T_6$ of $H.P. = \frac{1}{61} \implies T_6$ of $A.P. = a + 5d = 61$ $(i)$
$T_{10}$ of $H.P. = \frac{1}{105} \implies T_{10}$ of $A.P. = a + 9d = 105$ $(ii)$
Subtracting equation $(i)$ from $(ii):$
$(a + 9d) - (a + 5d) = 105 - 61$
$4d = 44 \implies d = 11$
Substituting $d = 11$ into equation $(i):$
$a + 5(11) = 61$
$a + 55 = 61 \implies a = 6$
Thus,the first term of the $H.P.$ is $\frac{1}{a} = \frac{1}{6}.$

(B) Let $a$ be the first term and $d$ be the common difference of the corresponding $A.P.$
The $p^{th}$ term of the $A.P.$ is $T_p = a + (p - 1)d = \frac{1}{q} \dots (i)$
The $q^{th}$ term of the $A.P.$ is $T_q = a + (q - 1)d = \frac{1}{p} \dots (ii)$
Subtracting $(ii)$ from $(i)$:
$(p - q)d = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq}$
$\Rightarrow d = \frac{1}{pq}$
Substituting $d$ into $(i)$:
$a + (p - 1)\frac{1}{pq} = \frac{1}{q}$
$a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q}$
$a = \frac{1}{pq}$
The $pq^{th}$ term of the $A.P.$ is $T_{pq} = a + (pq - 1)d = \frac{1}{pq} + (pq - 1)\frac{1}{pq} = \frac{1 + pq - 1}{pq} = 1$.
Since the $pq^{th}$ term of the $A.P.$ is $1$,the $pq^{th}$ term of the $H.P.$ is the reciprocal of $1$,which is $1$.

(B) If a sequence is in $H.P.$,then its reciprocal is in $A.P.$
Let the $A.P.$ be $a, a+d, a+2d, \dots$
The $4^{th}$ term of $H.P.$ is $\frac{3}{5}$,so the $4^{th}$ term of $A.P.$ is $\frac{5}{3}$. Thus,$a + 3d = \frac{5}{3}$.
The $8^{th}$ term of $H.P.$ is $\frac{1}{3}$,so the $8^{th}$ term of $A.P.$ is $3$. Thus,$a + 7d = 3$.
Subtracting the first equation from the second: $(a + 7d) - (a + 3d) = 3 - \frac{5}{3} \implies 4d = \frac{4}{3} \implies d = \frac{1}{3}$.
Substituting $d$ in the first equation: $a + 3(\frac{1}{3}) = \frac{5}{3} \implies a + 1 = \frac{5}{3} \implies a = \frac{2}{3}$.
The $6^{th}$ term of the $A.P.$ is $a + 5d = \frac{2}{3} + 5(\frac{1}{3}) = \frac{2}{3} + \frac{5}{3} = \frac{7}{3}$.
Therefore,the $6^{th}$ term of the $H.P.$ is the reciprocal of $\frac{7}{3}$,which is $\frac{3}{7}$.

(A) The harmonic mean $H$ between two numbers $p$ and $q$ is given by the formula:
$H = \frac{2pq}{p + q}$
We need to find the value of the expression $\frac{H}{p} + \frac{H}{q}$.
Substituting the value of $H$ into the expression:
$\frac{H}{p} + \frac{H}{q} = \frac{1}{p} \left( \frac{2pq}{p + q} \right) + \frac{1}{q} \left( \frac{2pq}{p + q} \right)$
Simplifying the terms:
$= \frac{2q}{p + q} + \frac{2p}{p + q}$
Combining the fractions since they have a common denominator:
$= \frac{2q + 2p}{p + q}$
$= \frac{2(p + q)}{p + q}$
Canceling the common term $(p + q)$:
$= 2$
Thus,the correct option is $A$.

(C) Given that $H$ is the harmonic mean between $a$ and $b$,we have $H = \frac{2ab}{a + b}$.
We need to evaluate the expression $\frac{1}{H - a} + \frac{1}{H - b}$.
Substituting the value of $H$:
$\frac{1}{\frac{2ab}{a + b} - a} + \frac{1}{\frac{2ab}{a + b} - b} = \frac{1}{\frac{2ab - a(a + b)}{a + b}} + \frac{1}{\frac{2ab - b(a + b)}{a + b}}$
$= \frac{a + b}{2ab - a^2 - ab} + \frac{a + b}{2ab - ab - b^2} = \frac{a + b}{ab - a^2} + \frac{a + b}{ab - b^2}$
$= \frac{a + b}{-a(a - b)} + \frac{a + b}{b(a - b)} = \frac{a + b}{a - b} \left( \frac{1}{b} - \frac{1}{a} \right)$
$= \frac{a + b}{a - b} \left( \frac{a - b}{ab} \right) = \frac{a + b}{ab} = \frac{a}{ab} + \frac{b}{ab} = \frac{1}{b} + \frac{1}{a}$.
Thus,the value is $\frac{1}{a} + \frac{1}{b}$.

(D) Let the roots of the quadratic equation $x^2 - 10x + 11 = 0$ be $\alpha$ and $\beta$.
From the properties of quadratic equations,the sum of the roots is $\alpha + \beta = -(-10)/1 = 10$.
The product of the roots is $\alpha \beta = 11/1 = 11$.
The Harmonic Mean $(H.M.)$ of two numbers $\alpha$ and $\beta$ is given by the formula $H.M. = \frac{2\alpha \beta}{\alpha + \beta}$.
Substituting the values,we get $H.M. = \frac{2 \times 11}{10} = \frac{22}{10} = \frac{11}{5}$.

(C) The harmonic mean $(H.M.)$ of two numbers $x$ and $y$ is given by the formula $H.M. = \frac{2xy}{x + y}$.
Here,$x = \frac{a}{1 - ab}$ and $y = \frac{a}{1 + ab}$.
First,calculate the product $xy = \left(\frac{a}{1 - ab}\right) \left(\frac{a}{1 + ab}\right) = \frac{a^2}{1 - a^2b^2}$.
Next,calculate the sum $x + y = \frac{a}{1 - ab} + \frac{a}{1 + ab} = \frac{a(1 + ab) + a(1 - ab)}{(1 - ab)(1 + ab)} = \frac{a + a^2b + a - a^2b}{1 - a^2b^2} = \frac{2a}{1 - a^2b^2}$.
Now,substitute these into the $H.M.$ formula:
$H.M. = \frac{2 \left(\frac{a^2}{1 - a^2b^2}\right)}{\frac{2a}{1 - a^2b^2}} = \frac{2a^2}{1 - a^2b^2} \times \frac{1 - a^2b^2}{2a} = \frac{2a^2}{2a} = a$.

(A) The formula for the $n^{th}$ Harmonic Mean $(H.M.)$ between two numbers $a$ and $b$ is given by $H_n = \frac{(n+1)ab}{na+b}$.
Here,$a = 3$,$b = \frac{6}{13}$,and $n = 6$.
Substituting these values into the formula:
$H_6 = \frac{(6+1) \times 3 \times \frac{6}{13}}{6 \times 3 + \frac{6}{13}}$
$H_6 = \frac{7 \times 3 \times \frac{6}{13}}{18 + \frac{6}{13}}$
$H_6 = \frac{\frac{126}{13}}{\frac{234+6}{13}}$
$H_6 = \frac{126}{240}$
Simplifying the fraction by dividing both numerator and denominator by $2$,we get:
$H_6 = \frac{63}{120}$.

(B) The harmonic mean $(HM)$ of two numbers $a$ and $b$ is given by $\frac{2ab}{a + b}$.
Given that $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n} = \frac{2ab}{a + b}$.
Cross-multiplying,we get $(a^{n + 1} + b^{n + 1})(a + b) = 2ab(a^n + b^n)$.
Expanding both sides: $a^{n + 2} + a^{n + 1}b + ab^{n + 1} + b^{n + 2} = 2a^{n + 1}b + 2ab^{n + 1}$.
Rearranging the terms: $a^{n + 2} + b^{n + 2} = a^{n + 1}b + ab^{n + 1}$.
$a^{n + 1}(a - b) - b^{n + 1}(a - b) = 0$.
$(a^{n + 1} - b^{n + 1})(a - b) = 0$.
Since $a \neq b$,we must have $a^{n + 1} = b^{n + 1}$.
This implies $(\frac{a}{b})^{n + 1} = 1 = (\frac{a}{b})^0$.
Therefore,$n + 1 = 0$,which gives $n = -1$.

(B) The harmonic mean $H$ between $a$ and $b$ is given by $H = \frac{2ab}{a + b}$.
We need to evaluate the expression $\frac{H + a}{H - a} + \frac{H + b}{H - b}$.
First,simplify the expression:
$\frac{H + a}{H - a} + \frac{H + b}{H - b} = \frac{(H + a)(H - b) + (H + b)(H - a)}{(H - a)(H - b)}$
$= \frac{(H^2 - Hb + aH - ab) + (H^2 - Ha + bH - ab)}{H^2 - Hb - aH + ab}$
$= \frac{2H^2 - 2ab}{H^2 - H(a + b) + ab}$.
Since $H = \frac{2ab}{a + b}$,we have $H(a + b) = 2ab$.
Substituting this into the denominator:
Denominator $= H^2 - 2ab + ab = H^2 - ab$.
Thus,the expression becomes $\frac{2(H^2 - ab)}{H^2 - ab} = 2$.

(C) Given that $a, b, c$ are in $H.P.$ (Harmonic Progression).
By definition,$b = \frac{2ac}{a+c}$.
We know that for any two positive numbers $a$ and $c$,the Arithmetic Mean $(A.M.)$ is greater than the Harmonic Mean $(H.M.)$.
$A.M. = \frac{a+c}{2}$ and $H.M. = b$.
So,$\frac{a+c}{2} > b$.
Using the Power Mean Inequality,for $n=2$,the quadratic mean is greater than the harmonic mean.
Alternatively,consider the expression $a^2 + c^2 - 2b^2$.
Substituting $b = \frac{2ac}{a+c}$,we get $a^2 + c^2 - 2(\frac{2ac}{a+c})^2 = a^2 + c^2 - \frac{8a^2c^2}{(a+c)^2}$.
$= \frac{(a^2+c^2)(a+c)^2 - 8a^2c^2}{(a+c)^2} = \frac{(a^2+c^2)(a^2+c^2+2ac) - 8a^2c^2}{(a+c)^2}$.
$= \frac{(a^2+c^2)^2 + 2ac(a^2+c^2) - 8a^2c^2}{(a+c)^2} = \frac{(a^2-c^2)^2 + 2ac(a^2+c^2-2ac)}{(a+c)^2} = \frac{(a^2-c^2)^2 + 2ac(a-c)^2}{(a+c)^2}$.
Since this expression is always positive for $a \neq c$,we have $a^2 + c^2 > 2b^2$.

(C) Given that $a, b, c, d$ are in $H.P.$
Since $a, b, c$ are in $H.P.$,$b$ is the $H.M.$ between $a$ and $c$. We know that for any two positive numbers,$A.M. > G.M. > H.M.$
For $a$ and $c$,$A.M. = \frac{a+c}{2}$,$G.M. = \sqrt{ac}$,and $H.M. = b$.
Since $A.M. > H.M.$,we have $\frac{a+c}{2} > b \Rightarrow a + c > 2b$ .... $(i)$
Since $G.M. > H.M.$,we have $\sqrt{ac} > b \Rightarrow ac > b^2$ .... $(ii)$
Similarly,for $b, c, d$ in $H.P.$,$c$ is the $H.M.$ between $b$ and $d$.
Thus,$A.M. > H.M. \Rightarrow \frac{b+d}{2} > c \Rightarrow b + d > 2c$ .... $(iii)$
And $G.M. > H.M. \Rightarrow \sqrt{bd} > c \Rightarrow bd > c^2$ .... $(iv)$
Adding $(i)$ and $(iii)$,we get $(a + c) + (b + d) > 2b + 2c \Rightarrow a + d > b + c$.
Multiplying $(ii)$ and $(iv)$,we get $(ac)(bd) > (b^2)(c^2) \Rightarrow abcd > b^2c^2 \Rightarrow ad > bc$.
Therefore,both $(a)$ and $(b)$ are correct.

(D) The $k$-th term of the series is given by $T_k = \frac{1}{1 + 2 + 3 + \dots + k} = \frac{1}{\frac{k(k + 1)}{2}} = \frac{2}{k(k + 1)}$.
Using partial fractions,we can write $T_k = 2 \left[ \frac{1}{k} - \frac{1}{k + 1} \right]$.
The sum of $(n + 1)$ terms is $S_{n + 1} = \sum_{k = 1}^{n + 1} T_k = \sum_{k = 1}^{n + 1} 2 \left[ \frac{1}{k} - \frac{1}{k + 1} \right]$.
Expanding the summation: $S_{n + 1} = 2 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n + 1} - \frac{1}{n + 2}) \right]$.
This is a telescoping series where intermediate terms cancel out,leaving $S_{n + 1} = 2 \left[ 1 - \frac{1}{n + 2} \right]$.
Simplifying the expression: $S_{n + 1} = 2 \left[ \frac{n + 2 - 1}{n + 2} \right] = \frac{2(n + 1)}{n + 2}$.

(C) Let $T_k$ be the $k^{th}$ term of the series.
The $k^{th}$ term is the sum of the first $k$ odd numbers: $T_k = 1 + 3 + 5 + \dots + (2k - 1) = k^2$.
We need to find the sum of $(n - 1)$ terms,which is $S_{n-1} = \sum_{k=1}^{n-1} T_k = \sum_{k=1}^{n-1} k^2$.
Using the formula for the sum of squares of the first $m$ natural numbers,$\sum_{k=1}^{m} k^2 = \frac{m(m + 1)(2m + 1)}{6}$.
Substituting $m = n - 1$:
$S_{n-1} = \frac{(n - 1)((n - 1) + 1)(2(n - 1) + 1)}{6} = \frac{(n - 1)(n)(2n - 2 + 1)}{6} = \frac{n(n - 1)(2n - 1)}{6}$.

(B) The series is $1^2, 2.2^2, 3^2, 2.4^2, 5^2, 2.6^2, \dots$
When $n$ is odd,the $n^{th}$ term is $n^2$.
Since $n$ is odd,$n-1$ is even.
The sum of the first $n-1$ terms is obtained by substituting $n-1$ for $n$ in the given formula for even terms:
$S_{n-1} = \frac{(n-1)((n-1) + 1)^2}{2} = \frac{(n-1)n^2}{2}$.
The sum of $n$ terms is $S_n = S_{n-1} + T_n$.
$S_n = \frac{(n-1)n^2}{2} + n^2 = n^2 \left( \frac{n-1}{2} + 1 \right) = n^2 \left( \frac{n-1+2}{2} \right) = \frac{n^2(n+1)}{2}$.
Verification: For $n=1$,$S_1 = 1^2 = 1$. Option $(b)$ gives $\frac{1^2(1+1)}{2} = 1$. For $n=3$,$S_3 = 1^2 + 2.2^2 + 3^2 = 1 + 8 + 9 = 18$. Option $(b)$ gives $\frac{3^2(3+1)}{2} = \frac{9 \times 4}{2} = 18$.

(B) The given series is $2^2 + 4^2 + 6^2 + \dots + (2n)^2$.
This can be written as $2^2(1^2 + 2^2 + 3^2 + \dots + n^2)$.
We know that the sum of the squares of the first $n$ natural numbers is given by $\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.
Therefore,the sum of the series is $4 \times \frac{n(n + 1)(2n + 1)}{6}$.
Simplifying this,we get $\frac{2n(n + 1)(2n + 1)}{3}$.

(D) The $n^{th}$ term of the series is given by the sum of the first $n$ natural numbers:
$T_n = 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}$
The sum of $n$ terms $S_n$ is the sum of $T_n$ from $k=1$ to $n$:
$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k(k + 1)}{2} = \frac{1}{2} \left[ \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \right]$
Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$:
$S_n = \frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right]$
Factor out $\frac{n(n+1)}{2}$:
$S_n = \frac{1}{2} \cdot \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 1 \right] = \frac{n(n+1)}{4} \left[ \frac{2n+1+3}{3} \right]$
$S_n = \frac{n(n+1)}{4} \cdot \frac{2n+4}{3} = \frac{n(n+1) \cdot 2(n+2)}{12} = \frac{n(n+1)(n+2)}{6}$.

(A) Given the $n^{th}$ term is $T_n = n(n + 1) = n^2 + n$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k)$.
Using the standard summation formulas $\sum k^2 = \frac{n(n + 1)(2n + 1)}{6}$ and $\sum k = \frac{n(n + 1)}{2}$,we get:
$S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}$.
Factor out $\frac{n(n + 1)}{2}$:
$S_n = \frac{n(n + 1)}{2} \left( \frac{2n + 1}{3} + 1 \right)$.
$S_n = \frac{n(n + 1)}{2} \left( \frac{2n + 1 + 3}{3} \right) = \frac{n(n + 1)(2n + 4)}{6}$.
$S_n = \frac{n(n + 1) \cdot 2(n + 2)}{6} = \frac{n(n + 1)(n + 2)}{3}$.

(C) Let the sum be $S_n = 1(1!) + 2(2!) + 3(3!) + \dots + n(n!)$.
We can rewrite the $k$-th term as $k(k!) = ((k + 1) - 1)k! = (k + 1)! - k!$.
Now,we sum these terms from $k = 1$ to $n$:
$S_n = \sum_{k=1}^{n} ((k + 1)! - k!)$
$S_n = (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + ((n + 1)! - n!)$
This is a telescoping series where intermediate terms cancel out.
$S_n = (n + 1)! - 1!$
$S_n = (n + 1)! - 1$.

(B) The given series is $S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)}$.
Using the method of partial fractions,we can write $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.
Substituting this into the sum,we get:
$S_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{n} - \frac{1}{n+1} \right)$.
This is a telescoping series where all intermediate terms cancel out:
$S_n = 1 - \frac{1}{n+1}$.
Simplifying the expression,we get $S_n = \frac{n+1-1}{n+1} = \frac{n}{n+1}$.

(B) The given series is $S = 3 \cdot 6 + 4 \cdot 7 + 5 \cdot 8 + \dots$ up to $(n - 2)$ terms.
Let the $k$-th term of the series be $T_k = (k + 2)(k + 5) = k^2 + 7k + 10$.
The sum of $m$ terms is $\sum_{k=1}^{m} (k^2 + 7k + 10)$,where $m = n - 2$.
Sum $= \sum_{k=1}^{m} k^2 + 7 \sum_{k=1}^{m} k + \sum_{k=1}^{m} 10 = \frac{m(m+1)(2m+1)}{6} + \frac{7m(m+1)}{2} + 10m$.
Substituting $m = n - 2$:
Sum $= \frac{(n-2)(n-1)(2n-3)}{6} + \frac{7(n-2)(n-1)}{2} + 10(n-2)$.
$= \frac{(n-2)}{6} [(n-1)(2n-3) + 21(n-1) + 60] = \frac{(n-2)}{6} [2n^2 - 5n + 3 + 21n - 21 + 60] = \frac{(n-2)(2n^2 + 16n + 42)}{6} = \frac{2n^3 + 12n^2 + 10n - 84}{6}$.

(A) Given the summation $\sum_{i = 1}^n (3i - 2)$.
Using the linearity property of summation,we can write:
$\sum_{i = 1}^n (3i - 2) = 3 \sum_{i = 1}^n i - \sum_{i = 1}^n 2$.
Since $\sum_{i = 1}^n i = \frac{n(n + 1)}{2}$ and $\sum_{i = 1}^n 2 = 2n$,we substitute these values:
$= 3 \left[ \frac{n(n + 1)}{2} \right] - 2n$.
$= \frac{3n^2 + 3n}{2} - 2n$.
$= \frac{3n^2 + 3n - 4n}{2}$.
$= \frac{3n^2 - n}{2} = \frac{n(3n - 1)}{2}$.

(B) The $n$-th term of the series is $T_n = n^2(n + 1) = n^3 + n^2$.
The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^3 + k^2) = \sum_{k=1}^n k^3 + \sum_{k=1}^n k^2$.
Using the standard summation formulas $\sum k^3 = [\frac{n(n+1)}{2}]^2$ and $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$:
$S_n = [\frac{n(n+1)}{2}]^2 + \frac{n(n+1)(2n+1)}{6}$
$S_n = \frac{n(n+1)}{2} [\frac{n(n+1)}{2} + \frac{2n+1}{3}]$
$S_n = \frac{n(n+1)}{2} [\frac{3n^2 + 3n + 4n + 2}{6}]$
$S_n = \frac{n(n+1)(3n^2 + 7n + 2)}{12}$.

(C) The $n^{th}$ term of the series is given by $T_n = n(n + 1)(n + 2)$.
Expanding this,we get $T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^3 + 3k^2 + 2k)$.
Using the standard summation formulas:
$\sum k^3 = \left[ \frac{n(n+1)}{2} \right]^2$,$\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum k = \frac{n(n+1)}{2}$.
$S_n = \left[ \frac{n(n+1)}{2} \right]^2 + 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$.
$S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1)$.
Taking $\frac{n(n+1)}{4}$ as a common factor:
$S_n = \frac{n(n+1)}{4} [n(n+1) + 2(2n+1) + 4]$.
$S_n = \frac{n(n+1)}{4} [n^2 + n + 4n + 2 + 4] = \frac{n(n+1)}{4} [n^2 + 5n + 6]$.
Since $n^2 + 5n + 6 = (n+2)(n+3)$,we get $S_n = \frac{1}{4}n(n + 1)(n + 2)(n + 3)$.

(C) The formula for the sum of the cubes of the first $n$ natural numbers is given by:
$S_n = \left[ \frac{n(n+1)}{2} \right]^2$
Here,$n = 15$.
Substituting the value of $n$ into the formula:
$S_{15} = \left[ \frac{15(15+1)}{2} \right]^2$
$S_{15} = \left[ \frac{15 \times 16}{2} \right]^2$
$S_{15} = (15 \times 8)^2$
$S_{15} = (120)^2$
$S_{15} = 14400$.

(B) The sum of the squares of the first $n$ natural numbers is given by $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$.
The sum of the first $n$ natural numbers is given by $\sum n = \frac{n(n+1)}{2}$.
According to the problem,$\sum n^2 = \sum n + 330$.
Substituting the formulas: $\frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{2} + 330$.
Rearranging the terms: $\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} = 330$.
Taking $\frac{n(n+1)}{2}$ as a common factor: $\frac{n(n+1)}{2} \left[ \frac{2n+1}{3} - 1 \right] = 330$.
Simplifying the expression inside the bracket: $\frac{n(n+1)}{2} \left[ \frac{2n+1-3}{3} \right] = 330$.
$\frac{n(n+1)}{2} \cdot \frac{2(n-1)}{3} = 330$.
$\frac{n(n+1)(n-1)}{3} = 330$.
$n(n^2-1) = 990$.
$n^3 - n - 990 = 0$.
Testing values,if $n=10$,then $10^3 - 10 = 1000 - 10 = 990$. Thus,$n = 10$.

(D) The $r$-th term of the series is $T_r = \cot^{-1}(r^2 + r + 1)$.
Using the identity $\cot^{-1} x = \tan^{-1} \left( \frac{1}{x} \right)$,we have $T_r = \tan^{-1} \left( \frac{1}{r^2 + r + 1} \right)$.
We can rewrite the denominator as $1 + r(r+1)$,so $T_r = \tan^{-1} \left( \frac{(r+1) - r}{1 + r(r+1)} \right)$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$,we get $T_r = \tan^{-1}(r+1) - \tan^{-1} r$.
The sum of the first $n$ terms is $S_n = \sum_{r=1}^{n} (\tan^{-1}(r+1) - \tan^{-1} r)$.
This is a telescoping series: $S_n = (\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots + (\tan^{-1}(n+1) - \tan^{-1} n)$.
$S_n = \tan^{-1}(n+1) - \tan^{-1} 1$.
Using $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$,$S_n = \tan^{-1} \left( \frac{n+1-1}{1+(n+1)(1)} \right) = \tan^{-1} \left( \frac{n}{n+2} \right)$.
Since $\tan^{-1} x = \cot^{-1} (1/x)$,$S_n = \cot^{-1} \left( \frac{n+2}{n} \right)$.
Thus,all options are equivalent.

(B) The $n^{th}$ row contains $n$ terms. The last term of the $(n-1)^{th}$ row is the sum of the first $(n-1)$ natural numbers,which is $\frac{(n-1)n}{2}$.
Therefore,the first term of the $n^{th}$ row is $\frac{n(n-1)}{2} + 1 = \frac{n^2 - n + 2}{2}$.
The $n^{th}$ row is an Arithmetic Progression $(A.P.)$ with $n$ terms,first term $a = \frac{n^2 - n + 2}{2}$,and common difference $d = 1$.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values: $S_n = \frac{n}{2}[2(\frac{n^2 - n + 2}{2}) + (n-1)(1)]$.
$S_n = \frac{n}{2}[n^2 - n + 2 + n - 1] = \frac{n}{2}(n^2 + 1)$.

(A) The given series is $\frac{1}{1} + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$
The $n^{th}$ term of the series is given by the sum of the first $n$ natural numbers divided by $n$.
$T_n = \frac{1 + 2 + 3 + \dots + n}{n}$
Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}$,we get:
$T_n = \frac{\frac{n(n + 1)}{2}}{n}$
$T_n = \frac{n(n + 1)}{2n}$
$T_n = \frac{n + 1}{2}$
Thus,the correct option is $A$.

(A) We know that the square of the sum of the first $n$ natural numbers is given by:
${\left( \sum_{r=1}^{n} r \right)}^2 = \left( \frac{n(n+1)}{2} \right)^2 = \sum_{r=1}^{n} r^2 + 2 \sum_{1 \le s < t \le n} st$
Rearranging the terms to find the sum of products taken two at a time:
$2 \sum_{s < t} st = {\left( \frac{n(n+1)}{2} \right)}^2 - \sum_{r=1}^{n} r^2$
Substituting the formula for the sum of squares $\sum r^2 = \frac{n(n+1)(2n+1)}{6}$:
$2 \sum_{s < t} st = \frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}$
Taking $\frac{n(n+1)}{12}$ as a common factor:
$2 \sum_{s < t} st = \frac{n(n+1)}{12} [3n(n+1) - 2(2n+1)]$
$2 \sum_{s < t} st = \frac{n(n+1)}{12} [3n^2 + 3n - 4n - 2] = \frac{n(n+1)(3n^2 - n - 2)}{12}$
Factoring the quadratic expression $(3n^2 - n - 2) = (n-1)(3n+2)$:
$2 \sum_{s < t} st = \frac{n(n+1)(n-1)(3n+2)}{12}$
Dividing by $2$:
$\sum_{s < t} st = \frac{1}{24}n(n - 1)(n + 1)(3n + 2)$.

(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.
Expanding this, we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.
The sum of the first $20$ terms is $S_{20} = \sum_{n=1}^{20} (4n^3 + 4n^2 + n) = 4 \sum_{n=1}^{20} n^3 + 4 \sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} n$.
Using the standard summation formulas:
$\sum_{n=1}^{N} n^3 = \left[ \frac{N(N+1)}{2} \right]^2$, $\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$, and $\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$.
For $N = 20$:
$4 \sum_{n=1}^{20} n^3 = 4 \left[ \frac{20 \cdot 21}{2} \right]^2 = 4 \cdot (210)^2 = 4 \cdot 44100 = 176400$.
$4 \sum_{n=1}^{20} n^2 = 4 \cdot \frac{20 \cdot 21 \cdot 41}{6} = 4 \cdot 2870 = 11480$.
$\sum_{n=1}^{20} n = \frac{20 \cdot 21}{2} = 210$.
Adding these values: $S_{20} = 176400 + 11480 + 210 = 188090$.

(A) The sum of the first $n$ cubes is given by $\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2$.
The sum of the first $n$ squares is given by $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
Therefore,the ratio is $\frac{\left[ \frac{n(n+1)}{2} \right]^2}{\frac{n(n+1)(2n+1)}{6}} = \frac{n^2(n+1)^2}{4} \times \frac{6}{n(n+1)(2n+1)} = \frac{3n(n+1)}{2(2n+1)}$.
Substituting $n = 12$:
Ratio $= \frac{3 \times 12 \times (12+1)}{2 \times (2 \times 12 + 1)} = \frac{3 \times 12 \times 13}{2 \times 25} = \frac{3 \times 6 \times 13}{25} = \frac{234}{25}$.

(C) The $n^{th}$ term of the series is given by $T_n = \frac{2n + 1}{\sum_{k=1}^{n} k^2}$.
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}$.
Substituting this into the expression for $T_n$:
$T_n = \frac{2n + 1}{\frac{n(n + 1)(2n + 1)}{6}} = \frac{6}{n(n + 1)}$.
Using partial fractions,we can write $T_n = 6 \left( \frac{1}{n} - \frac{1}{n + 1} \right)$.
The sum of $n$ terms $S_n = \sum_{k=1}^{n} T_k = 6 \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k + 1} \right)$.
This is a telescoping series:
$S_n = 6 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ... + (\frac{1}{n} - \frac{1}{n + 1}) \right]$.
$S_n = 6 \left( 1 - \frac{1}{n + 1} \right) = 6 \left( \frac{n + 1 - 1}{n + 1} \right) = \frac{6n}{n + 1}$.

(A) The given series is $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots + n \text{ terms}$.
The $r$-th term $T_r$ is given by the product of the $r$-th terms of three arithmetic progressions: $(1, 2, 3, \dots)$,$(3, 5, 7, \dots)$,and $(5, 8, 11, \dots)$.
$T_r = r \cdot (2r + 1) \cdot (3r + 2) = r(6r^2 + 4r + 3r + 2) = 6r^3 + 7r^2 + 2r$.
The sum of $n$ terms is $S_n = \sum_{r=1}^{n} T_r = 6 \sum r^3 + 7 \sum r^2 + 2 \sum r$.
Using the standard summation formulas:
$S_n = 6 \left[ \frac{n(n+1)}{2} \right]^2 + 7 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 2 \left[ \frac{n(n+1)}{2} \right]$.
$S_n = \frac{6n^2(n+1)^2}{4} + \frac{7n(n+1)(2n+1)}{6} + n(n+1)$.
$S_n = \frac{n(n+1)}{6} [9n(n+1) + 7(2n+1) + 6]$.
$S_n = \frac{n(n+1)}{6} [9n^2 + 9n + 14n + 7 + 6] = \frac{n(n+1)(9n^2 + 23n + 13)}{6}$.

(D) The $n$-th term of the series is given by $T_n = \frac{3^n - 1}{3^n} = 1 - \left(\frac{1}{3}\right)^n$.
The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left(1 - \left(\frac{1}{3}\right)^k\right)$.
$S_n = \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} \left(\frac{1}{3}\right)^k$.
$S_n = n - \left[ \frac{\frac{1}{3}(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}} \right]$.
$S_n = n - \left[ \frac{\frac{1}{3}(1 - 3^{-n})}{\frac{2}{3}} \right]$.
$S_n = n - \frac{1}{2}(1 - 3^{-n}) = n - \frac{1}{2} + \frac{1}{2} \cdot 3^{-n} = n + \frac{1}{2}(3^{-n} - 1)$.

(C) The expression $\sum\limits_{m = 1}^n {{m^2}}$ represents the sum of the squares of the first $n$ natural numbers.
By the standard formula for the sum of squares of the first $n$ natural numbers,we have:
$\sum\limits_{m = 1}^n {{m^2}} = 1^2 + 2^2 + 3^2 + ... + n^2 = \frac{{n(n + 1)(2n + 1)}}{6}$.

(B) The $n^{th}$ term of the series is given by $T_n = n(n + 1) = n^2 + n$.
To find the sum of $n$ terms,we calculate $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 + k)$.
Using the standard summation formulas $\sum k^2 = \frac{n(n + 1)(2n + 1)}{6}$ and $\sum k = \frac{n(n + 1)}{2}$,we get:
$S_n = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}$.
Factoring out $\frac{n(n + 1)}{6}$,we get:
$S_n = \frac{n(n + 1)}{6} [(2n + 1) + 3] = \frac{n(n + 1)(2n + 4)}{6}$.
$S_n = \frac{n(n + 1) \cdot 2(n + 2)}{6} = \frac{1}{3}n(n + 1)(n + 2)$.

(B) The sum is given by $\sum_{n=11}^{20} n^3 = \sum_{n=1}^{20} n^3 - \sum_{n=1}^{10} n^3$.
Using the formula $\sum_{n=1}^{k} n^3 = [\frac{k(k+1)}{2}]^2$:
For $k=20$,$\sum_{n=1}^{20} n^3 = [\frac{20(21)}{2}]^2 = (210)^2 = 44100$.
For $k=10$,$\sum_{n=1}^{10} n^3 = [\frac{10(11)}{2}]^2 = (55)^2 = 3025$.
Therefore,the sum is $44100 - 3025 = 41075$.
Since the number $41075$ ends in $5$,it is an odd integer and is divisible by $5$.

(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.
Expanding this,we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.
The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (4k^3 + 4k^2 + k)$.
Using the standard summation formulas $\sum k^3 = \frac{n^2(n+1)^2}{4}$,$\sum k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum k = \frac{n(n+1)}{2}$:
$S_n = 4 \left[ \frac{n^2(n+1)^2}{4} \right] + 4 \left[ \frac{n(n+1)(2n+1)}{6} \right] + \frac{n(n+1)}{2}$.
$S_n = n^2(n+1)^2 + \frac{2n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$.
Taking $\frac{n(n+1)}{6}$ as a common factor:
$S_n = \frac{n(n+1)}{6} [6n(n+1) + 4(2n+1) + 3]$.
$S_n = \frac{n(n+1)}{6} [6n^2 + 6n + 8n + 4 + 3] = \frac{n(n+1)}{6} (6n^2 + 14n + 7)$.

(B) Given the $n^{th}$ term $T_n = 3 + n(n - 1) = n^2 - n + 3$.
The sum of $n$ terms $S_n$ is given by $\sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (k^2 - k + 3)$.
Using the standard summation formulas:
$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,and $\sum_{k=1}^{n} 3 = 3n$.
$S_n = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} + 3n$.
Taking $\frac{n}{6}$ as a common factor:
$S_n = \frac{n}{6} [(n+1)(2n+1) - 3(n+1) + 18]$.
$S_n = \frac{n}{6} [2n^2 + 3n + 1 - 3n - 3 + 18] = \frac{n}{6} [2n^2 + 16]$.
$S_n = \frac{2n(n^2 + 8)}{6} = \frac{n(n^2 + 8)}{3} = \frac{n^3 + 8n}{3}$.

(D) The given series is $S = \sum_{k=1}^{n} k(2n - (2k - 1))$.
Expanding the general term $T_k = k(2n - 2k + 1) = 2nk - 2k^2 + k$.
Summing from $k=1$ to $n$:
$S = \sum_{k=1}^{n} (2nk - 2k^2 + k) = 2n \sum_{k=1}^{n} k - 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$.
Using standard summation formulas:
$S = 2n \cdot \frac{n(n + 1)}{2} - 2 \cdot \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}$.
$S = n^2(n + 1) - \frac{n(n + 1)(2n + 1)}{3} + \frac{n(n + 1)}{2}$.
Factor out $\frac{n(n + 1)}{6}$:
$S = \frac{n(n + 1)}{6} [6n - 2(2n + 1) + 3]$.
$S = \frac{n(n + 1)}{6} [6n - 4n - 2 + 3] = \frac{n(n + 1)(2n + 1)}{6}$.

(C) Observe the first term of each group:
Group $1$: $1$
Group $2$: $2$
Group $3$: $5$
Group $4$: $10$
Group $5$: $17$
Let $a_n$ be the first term of the $n^{th}$ group.
The sequence of first terms is $1, 2, 5, 10, 17, \dots$
Calculate the differences between consecutive terms:
$2 - 1 = 1$
$5 - 2 = 3$
$10 - 5 = 5$
$17 - 10 = 7$
The differences are $1, 3, 5, 7, \dots$, which is an arithmetic progression with common difference $2$.
The $n^{th}$ term of this sequence is given by $a_n = a_1 + \sum_{k=1}^{n-1} (2k - 1) = 1 + (n-1)^2$.
For the $11^{th}$ group, $n = 11$:
$a_{11} = (11 - 1)^2 + 1 = 10^2 + 1 = 100 + 1 = 101$.

(C) The sum of the squares of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
To find the sum $11^2 + 12^2 + \dots + 20^2$, we calculate the sum of the first $20$ squares and subtract the sum of the first $10$ squares.
Sum of first $20$ squares: $\sum_{k=1}^{20} k^2 = \frac{20(20+1)(2 \times 20 + 1)}{6} = \frac{20 \times 21 \times 41}{6} = 10 \times 7 \times 41 = 2870$.
Sum of first $10$ squares: $\sum_{k=1}^{10} k^2 = \frac{10(10+1)(2 \times 10 + 1)}{6} = \frac{10 \times 11 \times 21}{6} = 5 \times 11 \times 7 = 385$.
Required sum $= 2870 - 385 = 2485$.

(B) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.
Using partial fractions,we can write the general term as $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
Thus,the sum of the first $N$ terms is $S_N = \sum_{n=1}^{N} \left( \frac{1}{n} - \frac{1}{n+1} \right)$.
Expanding this,we get $S_N = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{N} - \frac{1}{N+1})$.
All intermediate terms cancel out,leaving $S_N = 1 - \frac{1}{N+1}$.
Taking the limit as $N \to \infty$,we get $S = \lim_{N \to \infty} (1 - \frac{1}{N+1}) = 1 - 0 = 1$.

(C) The $n^{th}$ term of the series is given by $T_n = \frac{{\frac{n(n + 1)}{2 \cdot 2}}}{{1^3 + 2^3 + 3^3 + \dots + n^3}}$.
Using the formula for the sum of cubes,$\sum_{k=1}^n k^3 = \left( \frac{n(n+1)}{2} \right)^2$,we get:
$T_n = \frac{{\frac{n(n + 1)}{4}}}{{\left( \frac{n(n + 1)}{2} \right)^2}} = \frac{{\frac{n(n + 1)}{4}}}{{\frac{n^2(n + 1)^2}{4}}} = \frac{1}{n(n + 1)}$.
Using partial fractions,$T_n = \frac{1}{n} - \frac{1}{n + 1}$.
The sum of $n$ terms $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k + 1} \right)$.
Expanding the sum: $S_n = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{n} - \frac{1}{n + 1})$.
All intermediate terms cancel out,leaving $S_n = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$.

(A) The $n^{th}$ term of the series is given by $T_n = \frac{1 + 2 + 3 + \dots + n}{n}$.
Using the formula for the sum of the first $n$ natural numbers,$\sum n = \frac{n(n + 1)}{2}$,we get:
$T_n = \frac{n(n + 1)}{2n} = \frac{n + 1}{2}$.
Now,the sum $S$ of $n$ terms is $S = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k + 1}{2}$.
$S = \frac{1}{2} \left( \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \right)$.
$S = \frac{1}{2} \left( \frac{n(n + 1)}{2} + n \right)$.
$S = \frac{1}{2} \left( \frac{n^2 + n + 2n}{2} \right) = \frac{n^2 + 3n}{4} = \frac{n(n + 3)}{4}$.

(B) The given series is $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$
Observe the numerators: $2, 7, 15, 26, \dots$
Let $a_n$ be the numerator of the $n^{th}$ term. The differences between consecutive terms are $5, 8, 11, \dots$,which form an Arithmetic Progression $(AP)$ with first term $a = 5$ and common difference $d = 3$.
The $n^{th}$ term of the numerator sequence can be expressed as the sum of the first $n$ terms of this $AP$ plus the initial value:
$a_n = 2 + \sum_{k=0}^{n-1} (5 + 3k) = 2 + \frac{n}{2} [2(5) + (n-1)3] = 2 + \frac{n}{2} [10 + 3n - 3] = 2 + \frac{n(3n + 7)}{2} = \frac{4 + 3n^2 + 7n}{2}$.
Alternatively,looking at the pattern provided in the options:
$T_n = \frac{2 + 5 + 8 + \dots + (3n - 1)}{n!} = \frac{\frac{n}{2}[2(2) + (n-1)3]}{n!} = \frac{\frac{n}{2}[4 + 3n - 3]}{n!} = \frac{n(3n + 1)}{2(n!)}$.

(D) Given $t_n = \frac{1}{4}(n + 2)(n + 3)$.
We need to find the sum $S = \sum_{n=1}^{2003} \frac{1}{t_n}$.
Substituting the expression for $t_n$,we get $\frac{1}{t_n} = \frac{4}{(n+2)(n+3)}$.
Using partial fractions,$\frac{1}{(n+2)(n+3)} = \frac{1}{n+2} - \frac{1}{n+3}$.
Thus,$S = 4 \sum_{n=1}^{2003} \left( \frac{1}{n+2} - \frac{1}{n+3} \right)$.
This is a telescoping series: $S = 4 \left[ (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + \dots + (\frac{1}{2005} - \frac{1}{2006}) \right]$.
All intermediate terms cancel out,leaving $S = 4 \left( \frac{1}{3} - \frac{1}{2006} \right)$.
$S = 4 \left( \frac{2006 - 3}{3 \times 2006} \right) = 4 \times \frac{2003}{6018} = \frac{8012}{6018} = \frac{4006}{3009}$.

(A) Let $S = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}$.
We can split the sum into odd and even terms: $S = \sum_{n=1,3,5,\dots}^{\infty} \frac{1}{n^4} + \sum_{n=2,4,6,\dots}^{\infty} \frac{1}{n^4}$.
Let $S_{odd} = \sum_{n=1,3,5,\dots}^{\infty} \frac{1}{n^4}$.
The even terms can be written as $\sum_{k=1}^{\infty} \frac{1}{(2k)^4} = \frac{1}{2^4} \sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{1}{16} S$.
Thus,$S = S_{odd} + \frac{1}{16} S$.
$S_{odd} = S - \frac{1}{16} S = \frac{15}{16} S$.
Substituting $S = \frac{\pi^4}{90}$,we get $S_{odd} = \frac{15}{16} \times \frac{\pi^4}{90} = \frac{\pi^4}{16 \times 6} = \frac{\pi^4}{96}$.