Multinomial theorem, Terms free from radical sign in the expansion of (a1/p + b1/q), Problems regarding to three/four consecutive terms or coefficients Questions in English

(C) Let the four consecutive binomial coefficients be $a = {}^nC_{r-1}$,$b = {}^nC_r$,$c = {}^nC_{r+1}$,and $d = {}^nC_{r+2}$.
We know that $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$.
Then,$\frac{a+b}{a} = 1 + \frac{b}{a} = 1 + \frac{r}{n-r+1} = \frac{n+1}{n-r+1}$.
Similarly,$\frac{b+c}{b} = 1 + \frac{c}{b} = 1 + \frac{n-r}{r+1} = \frac{n+1}{r+1}$.
And $\frac{c+d}{c} = 1 + \frac{d}{c} = 1 + \frac{n-r-1}{r+2} = \frac{n+1}{r+2}$.
Since the numerators are constant $(n+1)$,the terms are of the form $\frac{k}{n-r+1}, \frac{k}{r+1}, \frac{k}{r+2}$.
However,the standard property states that for consecutive binomial coefficients,the reciprocals of these terms form an $A.P.$,hence the terms themselves are in $H.P.$

(D) The coefficients of $T_r, T_{r+1}, T_{r+2}$ in $(1+x)^{14}$ are $^{14}C_{r-1}, ^{14}C_r, ^{14}C_{r+1}$ respectively.
Since they are in $A.P.$,we have $2 \cdot ^{14}C_r = ^{14}C_{r-1} + ^{14}C_{r+1}$.
Dividing by $^{14}C_r$,we get $2 = \frac{^{14}C_{r-1}}{^{14}C_r} + \frac{^{14}C_{r+1}}{^{14}C_r}$.
Using the formula $\frac{^{n}C_k}{^{n}C_{k-1}} = \frac{n-k+1}{k}$,we have $\frac{^{14}C_{r-1}}{^{14}C_r} = \frac{r}{14-r+1} = \frac{r}{15-r}$ and $\frac{^{14}C_{r+1}}{^{14}C_r} = \frac{14-r}{r+1}$.
So,$2 = \frac{r}{15-r} + \frac{14-r}{r+1}$.
$2(15-r)(r+1) = r(r+1) + (14-r)(15-r)$.
$2(15r + 15 - r^2 - r) = r^2 + r + 210 - 14r - 15r + r^2$.
$2(-r^2 + 14r + 15) = 2r^2 - 28r + 210$.
$-2r^2 + 28r + 30 = 2r^2 - 28r + 210$.
$4r^2 - 56r + 180 = 0$.
$r^2 - 14r + 45 = 0$.
$(r-5)(r-9) = 0$.
Thus,$r = 5$ or $r = 9$. Since $9$ is the only option provided,$r = 9$.

(B) The general term in the expansion of $(5^{1/2} + 7^{1/8})^{1024}$ is given by $T_{r+1} = {}^{1024}C_r (5^{1/2})^{1024-r} (7^{1/8})^r$,where $0 \le r \le 1024$.
This simplifies to $T_{r+1} = {}^{1024}C_r \cdot 5^{(1024-r)/2} \cdot 7^{r/8}$.
For the term to be an integer,the exponents of $5$ and $7$ must be non-negative integers.
Thus,$(1024-r)/2$ must be an integer,which implies $r$ must be even.
Also,$r/8$ must be an integer,which implies $r$ must be a multiple of $8$.
Since $r$ must be a multiple of $8$,it is automatically even.
Therefore,$r$ must be of the form $8k$,where $k$ is an integer such that $0 \le 8k \le 1024$.
This gives $0 \le k \le 128$.
The number of possible values for $k$ is $128 - 0 + 1 = 129$.
Thus,there are $129$ integral terms.

(B) In the expansion of $(y^{1/5} + x^{1/10})^{55}$,the general term is given by:
$T_{r+1} = {}^{55}C_r (y^{1/5})^{55-r} (x^{1/10})^r = {}^{55}C_r \cdot y^{11 - r/5} \cdot x^{r/10}$.
For the term to be free from radical signs,the exponents of $x$ and $y$ must be integers.
This requires $r/5$ and $r/10$ to be integers.
Since $0 \le r \le 55$,$r$ must be a multiple of $10$ (the least common multiple of $5$ and $10$).
The possible values for $r$ are $0, 10, 20, 30, 40, 50$.
There are $6$ such values,corresponding to the terms $T_1, T_{11}, T_{21}, T_{31}, T_{41}, T_{51}$.
Thus,there are $6$ terms free from radical signs.

(C) Let ${a_1}, {a_2}, {a_3}, {a_4}$ be the coefficients of the $(r+1)^{th}, (r+2)^{th}, (r+3)^{th}$ and $(r+4)^{th}$ terms in the expansion of ${(1 + x)^n}$ respectively.
Then ${a_1} = {^nC_r}, {a_2} = {^nC_{r+1}}, {a_3} = {^nC_{r+2}}, {a_4} = {^nC_{r+3}}$.
Using the identity ${^nC_r} + {^nC_{r+1}} = {^{n+1}C_{r+1}}$,we have:
$\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}} = \frac{{{^nC_r}}}{{{^{n+1}C_{r+1}}}} + \frac{{{^nC_{r+2}}}}{{{^{n+1}C_{r+3}}}}$
Using the property ${^{n+1}C_{k+1}} = \frac{n+1}{k+1} {^nC_k}$,we get:
$= \frac{{{^nC_r}}}{{\frac{n+1}{r+1} {^nC_r}}} + \frac{{{^nC_{r+2}}}}{{\frac{n+1}{r+3} {^nC_{r+2}}}} = \frac{r+1}{n+1} + \frac{r+3}{n+1} = \frac{2r+4}{n+1} = \frac{2(r+2)}{n+1}$.
Now,consider $\frac{2{a_2}}{{a_2} + {a_3}} = \frac{2{^nC_{r+1}}}{{^nC_{r+1}} + {^nC_{r+2}}} = \frac{2{^nC_{r+1}}}{{^{n+1}C_{r+2}}} = 2 \cdot \frac{{^nC_{r+1}}}{{\frac{n+1}{r+2} {^nC_{r+1}}}} = \frac{2(r+2)}{n+1}$.
Thus,$\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}} = \frac{{2{a_2}}}{{{a_2} + {a_3}}}$.

(A) In the expansion of $(1 + x)^n$,the coefficients of the second,third,and fourth terms are $^nC_1, ^nC_2$,and $^nC_3$ respectively.
Given that $^nC_1, ^nC_2, ^nC_3$ are in $A.P.$,we have:
$2(^nC_2) = ^nC_1 + ^nC_3$
$2 \times \frac{n(n - 1)}{2} = n + \frac{n(n - 1)(n - 2)}{6}$
Dividing by $n$ (since $n \neq 0$):
$n - 1 = 1 + \frac{(n - 1)(n - 2)}{6}$
$6(n - 1) = 6 + (n^2 - 3n + 2)$
$6n - 6 = n^2 - 3n + 8$
$n^2 - 9n + 14 = 0$
$(n - 7)(n - 2) = 0$
So,$n = 7$ or $n = 2$.
For $n = 2$,the expansion $(1 + x)^2$ has only three terms,so the fourth term does not exist. Thus,$n = 7$ is the only valid solution.

(A) The general term in the multinomial expansion of $(1 + x + y + z)^4$ is given by $\frac{4!}{n_1! n_2! n_3! n_4!} (1)^{n_1} (x)^{n_2} (y)^{n_3} (z)^{n_4}$,where $n_1 + n_2 + n_3 + n_4 = 4$.
$1$. For the coefficient of $x^2y$,we have $n_2=2, n_3=1, n_4=0$,so $n_1=1$. The coefficient is $\frac{4!}{1! 2! 1! 0!} = \frac{24}{2} = 12$.
$2$. For the coefficient of $xy^2z$,we have $n_2=1, n_3=2, n_4=1$,so $n_1=0$. The coefficient is $\frac{4!}{0! 1! 2! 1!} = \frac{24}{2} = 12$.
$3$. For the coefficient of $xyz$,we have $n_2=1, n_3=1, n_4=1$,so $n_1=1$. The coefficient is $\frac{4!}{1! 1! 1! 1!} = 24$.
The ratio is $12 : 12 : 24$,which simplifies to $1 : 1 : 2$.

(A) The general term of the expansion $(y^{1/5} + x^{1/10})^{55}$ is given by:
$T_{r+1} = ^{55}C_{r} (y^{1/5})^{55-r} (x^{1/10})^{r}$
$T_{r+1} = ^{55}C_{r} y^{\frac{55-r}{5}} x^{\frac{r}{10}}$
For the powers of $x$ and $y$ to be free from radical signs,the exponents $\frac{55-r}{5}$ and $\frac{r}{10}$ must be integers.
$\frac{55-r}{5} = 11 - \frac{r}{5}$ must be an integer,which implies $r$ must be a multiple of $5$.
$\frac{r}{10}$ must be an integer,which implies $r$ must be a multiple of $10$.
Combining these,$r$ must be a multiple of $10$ where $0 \le r \le 55$.
Possible values for $r$ are $0, 10, 20, 30, 40, 50$.
There are $6$ such values of $r$,so there are $6$ terms in the expansion that are free from radical signs.

(A) The general term in the expansion of $(a+2b+4ab)^{10}$ is given by the multinomial theorem as: $\frac{10!}{\alpha! \beta! \gamma!} a^{\alpha} (2b)^{\beta} (4ab)^{\gamma} = \frac{10!}{\alpha! \beta! \gamma!} a^{\alpha+\gamma} b^{\beta+\gamma} 2^{\beta} 4^{\gamma}$.
We are given the powers of $a$ and $b$ as $7$ and $8$ respectively,so:
$\alpha + \beta + \gamma = 10$ $(1)$
$\alpha + \gamma = 7$ $(2)$
$\beta + \gamma = 8$ $(3)$
Adding $(2)$ and $(3)$,we get $\alpha + \beta + 2\gamma = 15$. Subtracting $(1)$ from this,we get $\gamma = 5$.
Substituting $\gamma = 5$ into $(2)$ and $(3)$,we get $\alpha = 2$ and $\beta = 3$.
The coefficient is $\frac{10!}{2! 3! 5!} \cdot 2^{\beta} \cdot 4^{\gamma} = \frac{10!}{2! 3! 5!} \cdot 2^{3} \cdot (2^2)^{5} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{2 \cdot 6} \cdot 2^{3} \cdot 2^{10} = 2520 \cdot 2^{13} = 315 \cdot 8 \cdot 2^{13} = 315 \cdot 2^{16}$.
Thus,$K = 315$.

(B) Let the three consecutive coefficients be $^{n+2}C_{r-1}$,$^{n+2}C_{r}$,and $^{n+2}C_{r+1}$.
Given the ratio $^{n+2}C_{r-1} : ^{n+2}C_{r} : ^{n+2}C_{r+1} = 1 : 3 : 5$.
From $\frac{^{n+2}C_{r-1}}{^{n+2}C_{r}} = \frac{1}{3}$,we get $\frac{r}{n-r+3} = \frac{1}{3} \implies 3r = n-r+3 \implies n = 4r-3$ $(i)$.
From $\frac{^{n+2}C_{r}}{^{n+2}C_{r+1}} = \frac{3}{5}$,we get $\frac{r+1}{n-r+2} = \frac{3}{5} \implies 5r+5 = 3n-3r+6 \implies 3n = 8r-1$ $(ii)$.
Substituting $n = 4r-3$ into $(ii)$: $3(4r-3) = 8r-1 \implies 12r-9 = 8r-1 \implies 4r = 8 \implies r = 2$.
Then $n = 4(2)-3 = 5$.
The coefficients are $^{7}C_{1}, ^{7}C_{2}, ^{7}C_{3}$,which are $7, 21, 35$.
The sum is $7 + 21 + 35 = 63$.

(B) The general term in the expansion of $(1+2^{1/3}+3^{1/2})^6$ is given by $\frac{6!}{r_1! r_2! r_3!} (1)^{r_1} (2^{1/3})^{r_2} (3^{1/2})^{r_3}$,where $r_1+r_2+r_3=6$.
For the term to be rational,$r_2$ must be a multiple of $3$ and $r_3$ must be a multiple of $2$.
Possible values for $(r_1, r_2, r_3)$ are:
$1$. $(6, 0, 0) \implies \frac{6!}{6!0!0!} (1)^6 (2)^0 (3)^0 = 1$
$2$. $(4, 0, 2) \implies \frac{6!}{4!0!2!} (1)^4 (2)^0 (3)^1 = 15 \times 3 = 45$
$3$. $(2, 0, 4) \implies \frac{6!}{2!0!4!} (1)^2 (2)^0 (3)^2 = 15 \times 9 = 135$
$4$. $(0, 0, 6) \implies \frac{6!}{0!0!6!} (1)^0 (2)^0 (3)^3 = 1 \times 27 = 27$
$5$. $(3, 3, 0) \implies \frac{6!}{3!3!0!} (1)^3 (2)^1 (3)^0 = 20 \times 2 = 40$
$6$. $(1, 3, 2) \implies \frac{6!}{1!3!2!} (1)^1 (2)^1 (3)^1 = 60 \times 6 = 360$
$7$. $(0, 6, 0) \implies \frac{6!}{0!6!0!} (1)^0 (2)^2 (3)^0 = 1 \times 4 = 4$
Sum $= 1 + 45 + 135 + 27 + 40 + 360 + 4 = 612$.

(A) The coefficients of $T_r, T_{r+1}, T_{r+2}$ in $(a+b)^{12}$ are $^{12}C_{r-1}, ^{12}C_r, ^{12}C_{r+1}$.
Since they are in $G.P.$,we have $(^{12}C_r)^2 = (^{12}C_{r-1}) \times (^{12}C_{r+1})$.
Using the property $\frac{^{n}C_k}{^{n}C_{k-1}} = \frac{n-k+1}{k}$,we get $\frac{^{12}C_r}{^{12}C_{r-1}} = \frac{12-r+1}{r} = \frac{13-r}{r}$ and $\frac{^{12}C_{r+1}}{^{12}C_r} = \frac{12-r}{r+1}$.
Equating the ratios: $\frac{13-r}{r} = \frac{12-r}{r+1} \implies (13-r)(r+1) = r(12-r)$.
$13r + 13 - r^2 - r = 12r - r^2 \implies 12r + 13 = 12r \implies 13 = 0$,which is impossible.
Thus,there are no such values of $r$,so $p = 0$.
For the expansion $(3^{1/4} + 4^{1/3})^{12}$,the general term is $T_{k+1} = ^{12}C_k (3^{1/4})^{12-k} (4^{1/3})^k = ^{12}C_k \cdot 3^{(12-k)/4} \cdot 4^{k/3}$.
For the term to be rational,$(12-k)$ must be divisible by $4$ and $k$ must be divisible by $3$.
Possible values for $k \in \{0, 1, 2, \dots, 12\}$ are $k=0$ and $k=12$.
For $k=0$: $T_1 = ^{12}C_0 \cdot 3^3 \cdot 4^0 = 1 \cdot 27 \cdot 1 = 27$.
For $k=12$: $T_{13} = ^{12}C_{12} \cdot 3^0 \cdot 4^4 = 1 \cdot 1 \cdot 256 = 256$.
Sum $q = 27 + 256 = 283$.
Therefore,$p+q = 0 + 283 = 283$.

(C) Let the expression be $E = \left(\frac{x}{2} + \frac{1}{x} + \sqrt{2}\right)^5$.
We can rewrite the expression as:
$E = \left(\frac{x^2 + 2 + 2\sqrt{2}x}{2x}\right)^5 = \frac{(x + \sqrt{2})^{10}}{32x^5}$.
The constant term in the expansion of $E$ is the coefficient of $x^5$ in the expansion of $(x + \sqrt{2})^{10}$ divided by $32$.
Using the Binomial Theorem,the general term in $(x + \sqrt{2})^{10}$ is $T_{r+1} = {}^{10}C_r \cdot x^{10-r} \cdot (\sqrt{2})^r$.
For the coefficient of $x^5$,we set $10-r = 5$,which gives $r = 5$.
The term is ${}^{10}C_5 \cdot (\sqrt{2})^5 = 252 \cdot 4\sqrt{2} = 1008\sqrt{2}$.
Thus,the constant term is $\frac{1008\sqrt{2}}{32} = \frac{63\sqrt{2}}{2}$.
Comparing this with $\frac{a\sqrt{2}}{2}$,we get $a = 63$.

(A) The general term in the multinomial expansion of $(3+2x+x^2)^5$ is given by $\frac{5!}{p!q!r!} (3)^p (2x)^q (x^2)^r$,where $p+q+r=5$.
We need the coefficient of $x^5$,so $q+2r=5$.
Possible non-negative integer solutions for $(p, q, r)$ are:
$1$) If $r=0$,then $q=5$,$p=0$. Term: $\frac{5!}{0!5!0!} (3)^0 (2)^5 (1)^0 = 1 \times 1 \times 32 \times 1 = 32$.
$2$) If $r=1$,then $q=3$,$p=1$. Term: $\frac{5!}{1!3!1!} (3)^1 (2)^3 (1)^1 = 20 \times 3 \times 8 = 480$.
$3$) If $r=2$,then $q=1$,$p=2$. Term: $\frac{5!}{2!1!2!} (3)^2 (2)^1 (1)^2 = 30 \times 9 \times 2 = 540$.
Summing these coefficients: $32 + 480 + 540 = 1052$.

(A) The general term in the expansion of $(x-2y+3z)^6$ is given by the multinomial theorem as $\frac{6!}{a!b!c!} x^a (-2y)^b (3z)^c$,where $a+b+c=6$.
For the term $xy^2z^3$,we have $a=1, b=2, c=3$.
Substituting these values,the coefficient is $\frac{6!}{1! \times 2! \times 3!} \times (-2)^2 \times (3)^3$.
Calculating the values: $\frac{720}{1 \times 2 \times 6} \times 4 \times 27 = 60 \times 4 \times 27 = 240 \times 27 = 6480$.

(B) The coefficients of the $2nd$,$3rd$,and $4th$ terms in the expansion of $(1+x)^{n}$ are ${}^{n}C_{1}$,${}^{n}C_{2}$,and ${}^{n}C_{3}$ respectively.
Given that these are in arithmetic progression,we have $2({}^{n}C_{2}) = {}^{n}C_{1} + {}^{n}C_{3}$.
Substituting the values,$2 \times \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6}$.
Dividing by $n$ (since $n \neq 0$),we get $n-1 = 1 + \frac{(n-1)(n-2)}{6}$.
$6(n-2) = (n-1)(n-2)$.
Since $n \neq 2$,we divide by $(n-2)$ to get $6 = n-1$,which implies $n = 7$.
The sum of the coefficients of odd powers of $x$ in the expansion of $(1+x)^{n}$ is given by $2^{n-1}$.
Substituting $n=7$,the sum is $2^{7-1} = 2^{6} = 64$.