Question

Let $S_k=k=1,2,....100$ denote the sum of the infinite G.P. whose first term is $\frac{k-1}{k!}$ and the common ratio is $\frac{1}{k}$. Then the value of $\frac{{100}^2}{100!}+\sum _{k=1}^{100}|(k^2-3k+1)S_k|$ is

Answer 1

We know that $S_{\infty }$ of a G.P$=\frac{a}{1-r}$

Given$:S_k$ denotes the sum of infinite G.P. whose first term is $\frac{k-1}{k!}$ and common ratio is $\frac{1}{k}$

$\Rightarrow S_k=\frac{\frac{k-1}{k!}}{1-\frac{1}{k}}$

$=\frac{\frac{k-1}{k!}}{\frac{k-1}{k}}$

$=\frac{k-1}{k!}\times \frac{k}{k-1}$

$=\frac{k}{k(k-1)!}$

$=\frac{1}{(k-1)!}$

$\left|{\sum }_1^{100}(k^2-3k+1)S_k\right|$

$=\left|{\sum }_1^{100}(k^2-3k+1)\frac{1}{(k-1)!}\right|$

$=\left|{\sum }_1^{100}\frac{k^2-2k+1-k}{(k-1)!}\right|$

$=\left|{\sum }_1^{100}\frac{{(k-1)}^2-K}{(k-1)!}\right|$

$={\sum }_1^{100}|\frac{{(k-1)}^2}{(k-1)!}-\frac{k}{(k-1)!}|$

$={\sum }_1^{100}|\frac{k-1}{(k-1)!}-\frac{k}{(k-1)!}|$

$=0+|\frac{1}{0!}-\frac{2}{1!}|+|\frac{2}{1!}-\frac{3}{2!}|+|\frac{3}{2!}-\frac{4}{3!}|+.....+|\frac{99}{100!}-\frac{100}{99!}|$

$=1+\frac{2}{1!}-\frac{3}{2!}+\frac{3}{2!}-\frac{4}{3!}+.....+\frac{99}{100!}-\frac{100}{99!}$ and positive and negative terms having same value get cancelled

$=3-\frac{100}{99!}$

Again $\frac{{100}^2}{100!}+\left|{\sum }_1^{100}(k^2-3k+1)S_k\right|$

$=\frac{100\times 100}{100!}+3-\frac{100}{99!}$

$=\frac{100\times 100}{100\times 99!}+3-\frac{100}{99!}$

$=\frac{100}{99!}+3-\frac{100}{99!}=3$