91Ó°ÊÓ

Consider the given question,

$\underset{k=0}{\overset{\mathrm{∞}}{∑}} \left(\frac{2^k}{(k+3)!}−\frac{2^{k+1}}{(k+4)!}\right)$

The first term of the given series is obtained by substituting $k=0$,

$\begin{array}{r}=\frac{2^0}{(0+3)!}−\frac{2^{0+1}}{(0+4)!}\\ =\frac{1}{3!}−\frac{2}{4!}\\ =\frac{1}{6}−\frac{2}{4!}\end{array}$

First term is $\frac{1}{6}−\frac{2}{4!}$.

The second term of the given series is obtained by substituting $k=1$,

$\begin{array}{r}=\frac{2^k}{(k+3)!}−\frac{2^{k+1}}{(k+4)!}\\ =\frac{2^1}{(1+3)!}−\frac{2^{1+1}}{(1+4)!}\\ =\frac{2}{4!}−\frac{2^2}{5!}\end{array}$

Second term is$\frac{2}{4!}−\frac{2^2}{5!}$.

The third term of the given series is obtained by substituting $k=2$,

$\begin{array}{r}=\frac{2^k}{(k+3)!}−\frac{2^{k+1}}{(k+4)!}\\ =\frac{2^2}{(2+3)!}−\frac{2^{2+1}}{(2+4)!}\\ =\frac{2^2}{5!}−\frac{2^3}{6!}\end{array}$

Third term is $\frac{2^2}{5!}−\frac{2^3}{6!}$.

The fourth term of the given series is obtained by substituting $k=3$,

$\begin{array}{r}=\frac{2^k}{(k+3)!}−\frac{2^{k+1}}{(k+4)!}\\ =\frac{2^3}{(3+3)!}−\frac{2^{3+1}}{(3+4)!}\\ =\frac{2^3}{6!}−\frac{2^4}{7!}\end{array}$

The fifth term of the given series is obtained by substituting $k=4$,

$\begin{array}{r}=\frac{2^k}{(k+3)!}−\frac{2^{k+1}}{(k+4)!}\\ =\frac{2^4}{(4+3)!}−\frac{2^{4+1}}{(4+4)!}\\ =\frac{2^4}{7!}−\frac{2^5}{8!}\end{array}$

Fifth term is $\frac{2^4}{7!}−\frac{2^5}{8!}$.

The first and second terms in the sequence of partial sum is given below,

$\begin{array}{r}{S}_1=\frac{1}{6}−\frac{2}{4!}\\ =\frac{1}{6}\\ S_2=S_1+a_2\\ =\frac{1}{3!}−\frac{2}{4!}+\frac{2}{4!}−\frac{2^2}{5!}\\ =\frac{1}{3!}−\frac{2^2}{5!}\end{array}$

The third, fourth and fifth terms in the sequence of partial sum is given below,

$\begin{array}{r}{S}_3=S_2+a_3\\ =\frac{1}{3!}−\frac{2^2}{5!}+\frac{2^2}{5!}−\frac{2^3}{6!}\\ =\frac{1}{3!}−\frac{2^3}{6!}\\ S_4=S_3+a_4\\ =\frac{1}{3!}−\frac{2^3}{6!}+\frac{2^3}{6!}−\frac{2^4}{7!}\\ =\frac{1}{3!}−\frac{2^4}{7!}\\ S_5=S_4+a_5\\ =\frac{1}{3!}−\frac{2^4}{7!}+\frac{2^4}{7!}−\frac{2^5}{8!}\\ =\frac{1}{3!}−\frac{2^5}{8!}\end{array}$

The kth term in the sequence of the partial sums is given below,

$S_k=\left(\frac{1}{3!}−\frac{2}{4!}\right)+\left(\frac{2}{4!}−\frac{2^2}{5!}\right)+\left(\frac{2^2}{5!}−\frac{2^3}{6!}\right)+…+\left(\frac{2^k}{(k+3)!}−\frac{2^{k+1}}{(k+4)!}\right)$

In each two consecutive pairs, the second term of a pair cancels with the first term of the subsequent pair.

Thus, the series is telescopic.

The general term in its sequence of partial sums is $S_k=\frac{1}{3!}−\frac{2^{k+1}}{(k+4)!}$.

The $S_k$in its sequence of partial sums is $S_k=\frac{1}{3!}−\frac{2^{k+1}}{(k+4)!}$.

The value of $S_k$ is given below,

$\begin{array}{r}\underset{k→\mathrm{∞}}{lim} S_k=\underset{k→\mathrm{∞}}{lim} \left(\frac{1}{3!}−\frac{2^{k+1}}{(k+4)!}\right)\\ =\frac{1}{3!}−0\left(As\frac{b^k}{k!}→0,k→\mathrm{∞}\right)\\ =\frac{1}{6}\end{array}$