Question

Let $S_k$, where k 1,2, ...., 100, denote% the sum of infinite geometric series whose firm term is $\frac{k-1}{k!}$ and the common ratio is $\frac{1}{k}$. Then the value of $\frac{{100}^2}{100!}+{\sum }_{k=1}^{100}1(k^2-3k+1)S_k$ is ....

Answer 1

now question says $S_k$ is infinite sum whose first term is $a=\frac{k-1}{k!}$ and common ratio $r=\frac{1}{k}$
we know that sum of infinite GP $=\frac{a}{1-r}$ , now $S_k=\frac{(k-1)k}{k!(k-1)}$ $=\frac{k}{k!}=\frac{1}{(k-1)!}$

now from given question :$\frac{{100}^2}{100!}+{\Sigma }_{k=1}^{100}1(k^2-3k+1)S_k$ , now put the value of $s_k$

$=\frac{{100}^2}{100!}+{\Sigma }_{k=1}^{100}\frac{(k^2-3k+1)}{(k-1)!}$

$=\frac{{100}^2}{100!}+{\Sigma }_{k=1}^{100}\frac{(k-1{)}^2-k}{(k-1)!}$

$=\frac{{100}^2}{100!}+{\Sigma }_{k=1}^{100}\frac{(k-1{)}^2}{(k-1)!}-{\Sigma }_{k=1}^{100}\frac{k}{(k-1)!}$

now expanding summation series we get:

$=\frac{{100}^2}{100!}+[\frac{0^2}{0!}+\frac{1^2}{1!}+\frac{2^2}{2!}+............+\frac{{99}^2}{99!}]-[\frac{1}{0!}+\frac{2}{1!}+\frac{3}{2!}+...........+\frac{100}{99!}]$

$=\frac{100\times 100}{100\times 99!}+[\frac{0\times 0}{0!}+\frac{1\times 1}{1\times 0!}+\frac{2\times 2}{2\times 1!}+\frac{3\times 3}{3\times 2!}+.........+\frac{99\times 99}{99\times 98!}]-[\frac{1}{0!}+\frac{2}{1!}+\frac{3}{2!}+..........+\frac{99}{98!}]-\frac{100}{99!}$

$=\frac{100}{99!}+[\frac{1}{0!}+\frac{2}{1!}+\frac{3}{2!}+.........+\frac{99}{98!}]-[\frac{1}{0!}+\frac{2}{1!}+\frac{3}{2!}+.........+\frac{99}{98!}]-\frac{100}{99!}$

now each term will be cancle each other.

$=0$ $answer$