Exponential series Questions in English

(B) We have $\frac{1-2x}{e^x} = (1-2x)e^{-x}$.
Using the expansion $e^{-x} = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!} = \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!}$,we get:
$(1-2x) \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!} = \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!} - 2x \sum_{k=0}^{\infty} (-1)^k \frac{x^k}{k!}$.
The coefficient of $x^n$ is obtained from the first term when $k=n$ and from the second term when $k=n-1$:
Coefficient of $x^n = \frac{(-1)^n}{n!} - 2 \cdot \frac{(-1)^{n-1}}{(n-1)!}$.
Since $(-1)^{n-1} = -(-1)^n$,we have:
Coefficient of $x^n = \frac{(-1)^n}{n!} + 2 \cdot \frac{(-1)^n}{(n-1)!} = \frac{(-1)^n}{n!} [1 + 2n] = (-1)^n \cdot \frac{(1+2n)}{n!}$.

(B) We know that the expansion of $e^{-x}$ is given by $e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \dots + \frac{(-1)^n x^n}{n!} + \dots$
Now,consider the expression $(2+3x)e^{-x}$:
$(2+3x)e^{-x} = 2e^{-x} + 3xe^{-x}$
The term containing $x^{10}$ in $2e^{-x}$ is $2 \times \frac{(-x)^{10}}{10!} = \frac{2}{10!} x^{10}$.
The term containing $x^{10}$ in $3xe^{-x}$ is $3x \times \frac{(-x)^9}{9!} = 3x \times \frac{-x^9}{9!} = -\frac{3}{9!} x^{10}$.
To combine these,we write the second term with a denominator of $10!$:
$-\frac{3}{9!} = -\frac{3 \times 10}{10!} = -\frac{30}{10!}$.
Thus,the coefficient of $x^{10}$ is $\frac{2}{10!} - \frac{30}{10!} = \frac{2-30}{10!} = \frac{-28}{10!}$.

(B) Given the equation: $2 \cosh 2x + 10 \sinh 2x = 5$
Substitute the exponential definitions $\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}$ and $\sinh 2x = \frac{e^{2x} - e^{-2x}}{2}$:
$2 \left(\frac{e^{2x} + e^{-2x}}{2}\right) + 10 \left(\frac{e^{2x} - e^{-2x}}{2}\right) = 5$
$(e^{2x} + e^{-2x}) + 5(e^{2x} - e^{-2x}) = 5$
$e^{2x} + e^{-2x} + 5e^{2x} - 5e^{-2x} = 5$
$6e^{2x} - 4e^{-2x} = 5$
Multiply by $e^{2x}$:
$6(e^{2x})^2 - 5e^{2x} - 4 = 0$
Let $u = e^{2x}$,then $6u^2 - 5u - 4 = 0$
Factor the quadratic: $6u^2 - 8u + 3u - 4 = 0 \Rightarrow 2u(3u - 4) + 1(3u - 4) = 0$
$(2u + 1)(3u - 4) = 0$
Since $u = e^{2x} > 0$,we must have $3u - 4 = 0$,so $u = \frac{4}{3}$
$e^{2x} = \frac{4}{3} \Rightarrow 2x = \log \left(\frac{4}{3}\right)$
$x = \frac{1}{2} \log \left(\frac{4}{3}\right)$

(A) We have the series $\sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$.
Rewrite the numerator as $(2n+1) - 1$:
$\sum_{n=1}^{\infty} \frac{2n+1-1}{(2n+1)!} = \sum_{n=1}^{\infty} \left( \frac{2n+1}{(2n+1)!} - \frac{1}{(2n+1)!} \right) = \sum_{n=1}^{\infty} \left( \frac{1}{(2n)!} - \frac{1}{(2n+1)!} \right)$.
Expanding the summation:
$= \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{4!} - \frac{1}{5!} \right) + \dots$
Recall the Taylor series for $e^x$ at $x=1$: $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ and $e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$
Thus,$\frac{1}{e} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$
Therefore,the sum is equal to $\frac{1}{e}$.

(D) The general term of the series is $T_n = \frac{1+2+2^2+\ldots+2^{n-1}}{n !}$ for $n \geq 1$.
Using the sum of a geometric progression,$T_n = \frac{2^n-1}{n !}$.
We can rewrite this as $T_n = \frac{2^n}{n !} - \frac{1}{n !}$.
The sum of the series is $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{2^n}{n !} - \sum_{n=1}^{\infty} \frac{1}{n !}$.
Recall the expansion $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n !} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots$.
Thus,$\sum_{n=1}^{\infty} \frac{2^n}{n !} = e^2 - 1$ and $\sum_{n=1}^{\infty} \frac{1}{n !} = e - 1$.
Substituting these values,$S = (e^2 - 1) - (e - 1) = e^2 - e$.

(B) The given series is of the form $1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots = e^y$,where $y = x \log_e a$.
Substituting $y = \log_e a^x$,we get:
$1 + \log_e a^x + \frac{(\log_e a^x)^2}{2!} + \frac{(\log_e a^x)^3}{3!} + \ldots = e^{\log_e a^x}$.
Since $e^{\log_e z} = z$,we have $e^{\log_e a^x} = a^x$.

(A) Given,$x=1+\frac{1}{2 \times 1 !}+\frac{1}{4 \times 2 !}+\frac{1}{8 \times 3 !}+\ldots$
This is the expansion of $e^{1/2}$,so $x = e^{1/2}$.
Squaring both sides,we get $x^{2} = e$.
Now,$y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$
This is the expansion of $e^{x^{2}}$,so $y = e^{x^{2}}$.
Substituting $x^{2} = e$,we get $y = e^{e}$.
Taking $\log_{e}$ on both sides,$\log_{e} y = \log_{e} (e^{e}) = e \log_{e} e = e \times 1 = e$.

(B) Given,$f(x) = \sum_{n=1}^{\infty} \frac{2^n - 1}{n!} x^n$.
We can rewrite this as $f(x) = \sum_{n=1}^{\infty} \frac{(2x)^n}{n!} - \sum_{n=1}^{\infty} \frac{x^n}{n!}$.
Since $e^y = 1 + \sum_{n=1}^{\infty} \frac{y^n}{n!}$,we have $\sum_{n=1}^{\infty} \frac{y^n}{n!} = e^y - 1$.
Substituting this,we get $f(x) = (e^{2x} - 1) - (e^x - 1) = e^{2x} - e^x$.
Setting $f(x) = 0$,we have $e^{2x} - e^x = 0$,which implies $e^x(e^x - 1) = 0$.
Since $e^x > 0$ for all real $x$,we must have $e^x - 1 = 0$,which gives $e^x = 1$.
Thus,$x = 0$ is the only real solution.

(B) The general term of the series is $t_n = \frac{2n}{(2n+1)!}$.
We can rewrite this as $t_n = \frac{(2n+1) - 1}{(2n+1)!} = \frac{1}{(2n)!} - \frac{1}{(2n+1)!}$.
Summing from $n=1$ to $\infty$,we get $\sum_{n=1}^{\infty} \left( \frac{1}{(2n)!} - \frac{1}{(2n+1)!} \right) = \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots \right)$.
Recall the expansion $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$. For $x = -1$,$e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$.
Thus,the sum is $e^{-1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$.
Therefore,the value is $e^{-1}$.

(C) Let the given series be $S = 1 + \frac{1}{2!} + \frac{1 \cdot 3}{4!} + \frac{1 \cdot 3 \cdot 5}{6!} + \dots$
The general term $T_n$ for $n \ge 1$ is given by $T_n = \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{(2n)!}$.
We can rewrite the numerator as $\frac{(2n)!}{2^n n!}$.
Thus,$T_n = \frac{(2n)!}{2^n n! (2n)!} = \frac{1}{2^n n!}$.
The series is $S = \sum_{n=0}^{\infty} \frac{1}{2^n n!}$,where the $n=0$ term is $1$.
This is the expansion of $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ with $x = \frac{1}{2}$.
Therefore,$S = e^{1/2} = \sqrt{e}$.

(C) The general term $T_n$ of the series $1, 8, 21, 40, 65, \ldots$ in the numerator can be found using the method of differences. The first differences are $7, 13, 19, 25, \ldots$,which form an arithmetic progression with common difference $6$.
Thus,$T_n = an^2 + bn + c$.
For $n=1, T_1 = a+b+c = 1$.
For $n=2, T_2 = 4a+2b+c = 8$.
For $n=3, T_3 = 9a+3b+c = 21$.
Solving these,we get $a=3, b=-2, c=0$. So,$T_n = 3n^2 - 2n = n(3n-2)$.
The series is $S = \sum_{n=1}^{\infty} \frac{n(3n-2)}{n!} = \sum_{n=1}^{\infty} \frac{3n^2-2n}{n!}$.
Since the first term is $1$ (for $n=1$),we can write $S = \sum_{n=1}^{\infty} \frac{3n-2}{(n-1)!}$.
Let $k = n-1$,then $n = k+1$.
$S = \sum_{k=0}^{\infty} \frac{3(k+1)-2}{k!} = \sum_{k=0}^{\infty} \frac{3k+1}{k!} = 3\sum_{k=1}^{\infty} \frac{k}{k!} + \sum_{k=0}^{\infty} \frac{1}{k!} = 3\sum_{k=1}^{\infty} \frac{1}{(k-1)!} + e = 3e + e = 4e$.
Since $2 < e < 3$,we have $8 < 4e < 12$.
Therefore,$8 < S < 12$.

(C) The $n$-th term of the series is given by $t_n = \frac{\sum_{k=1}^{n+1} k^2}{(n+2)!}$.
Using the formula $\sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}$,we have $t_n = \frac{(n+1)(n+2)(2n+3)}{6(n+2)!} = \frac{(n+1)(2n+3)}{6(n+2)!} = \frac{2n^2+5n+3}{6(n+2)!}$.
Alternatively,let $r = n+1$,then $t_r = \frac{\sum_{k=1}^{r+1} k^2}{(r+2)!} = \frac{(r+1)(r+2)(2r+3)}{6(r+2)!} = \frac{(r+1)(2r+3)}{6(r+2)!} = \frac{2r^2+5r+3}{6(r+2)!}$.
Simplifying the expression: $t_r = \frac{2(r+2)(r+1) + r + 1 - 2}{6(r+2)!} = \frac{2}{6r!} + \frac{r+1}{6(r+2)!} = \frac{1}{3r!} + \frac{r+2-1}{6(r+2)!} = \frac{1}{3r!} + \frac{1}{6(r+1)!} - \frac{1}{6(r+2)!}$.
Summing from $r=1$ to $\infty$:
$S = \sum_{r=1}^{\infty} \left( \frac{2}{6r!} + \frac{3}{6(r+1)!} \right) = \frac{1}{3} \sum_{r=1}^{\infty} \frac{1}{r!} + \frac{1}{2} \sum_{r=1}^{\infty} \frac{1}{(r+1)!}$.
Since $e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2!} + \dots$,we have $\sum_{r=1}^{\infty} \frac{1}{r!} = e-1$ and $\sum_{r=1}^{\infty} \frac{1}{(r+1)!} = e - 1 - 1 = e-2$.
$S = \frac{1}{3}(e-1) + \frac{1}{2}(e-2) = \frac{2e-2+3e-6}{6} = \frac{5e-8}{6} = \frac{5e}{6} - \frac{4}{3}$.
Re-evaluating the original series sum: $t_n = \frac{2n+3}{6n!}$.
$S = \sum_{n=1}^{\infty} \frac{2n+3}{6n!} = \frac{1}{6} \sum_{n=1}^{\infty} (\frac{2}{(n-1)!} + \frac{3}{n!}) = \frac{1}{6} (2e + 3(e-1)) = \frac{5e-3}{6} = \frac{5e}{6} - \frac{1}{2}$.

(D) Given expression: $\frac{e^{7x}+e^x}{e^{3x}} = e^{4x} + e^{-2x}$.
Using the Taylor series expansion for $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$,we have:
$e^{4x} = \sum_{n=0}^{\infty} \frac{(4x)^n}{n!} = \sum_{n=0}^{\infty} \frac{4^n x^n}{n!}$.
$e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-2)^n x^n}{n!}$.
Adding these two series,the coefficient of $x^n$ is $\frac{4^n + (-2)^n}{n!}$.

(C) The sum of the first $(r-1)$ natural numbers is given by $\frac{(r-1)r}{2}$.
Substituting this into the summation,we get $\sum_{r=2}^{\infty} \frac{(r-1)r}{2 \cdot r!}$.
Since $r! = r(r-1)(r-2)!$,the expression simplifies to $\sum_{r=2}^{\infty} \frac{1}{2(r-2)!}$.
Factoring out the constant $\frac{1}{2}$,we have $\frac{1}{2} \sum_{r=2}^{\infty} \frac{1}{(r-2)!}$.
Let $k = r-2$. As $r$ goes from $2$ to $\infty$,$k$ goes from $0$ to $\infty$.
Thus,the expression becomes $\frac{1}{2} \sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{2} e$.

(B) We are given the series $S = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{k(k+1)}{k!}$.
Note that $\frac{k(k+1)}{k!} = \frac{k(k-1+2)}{k!} = \frac{k(k-1)}{k!} + \frac{2k}{k!} = \frac{1}{(k-2)!} + \frac{2}{(k-1)!}$ for $k \ge 2$.
For $k=1$,the term is $(-1)^{1+1}\frac{1(2)}{1!} = 2$.
Expanding the sum: $S = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{k(k-1)}{k!} + \sum_{k=1}^{\infty} (-1)^{k+1} \frac{2k}{k!}$.
$S = \sum_{k=2}^{\infty} \frac{(-1)^{k+1}}{(k-2)!} + 2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(k-1)!}$.
Let $j = k-2$ in the first sum and $m = k-1$ in the second sum:
$S = -\sum_{j=0}^{\infty} \frac{(-1)^{j}}{j!} + 2 \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!}$.
Since $\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = e^{-1} = \frac{1}{e}$,we have:
$S = -(\frac{1}{e}) + 2(\frac{1}{e}) = \frac{1}{e}$.