Vinod Singh and Shweta Pawar Solutions for Chapter: Sequence and Series, Exercise 2: Critical Thinking

$\frac{1^3+2^3+3^3+\dots +{12}^3}{1^2+2^2+3^2+\dots +{12}^2}$ is equal to

If $S_1, S_2$ and $S_3$ are the sum of first $n$ natural numbers, their squares and their cubes respectively, then  $S_3\left(1+8S_1\right)=$

$\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\dots$ equals

The sum of the series $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\dots .$ will be

Sum of series from $n=1$ to $n=\infty$ whose $n^{\mathrm{th}}$ term is $\frac{1}{\left(n+1\right)!}$ is

$\frac{{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{1}{4!}}+{\displaystyle \frac{1}{6!}}+\dots }{{\displaystyle \frac{1}{1!}}+{\displaystyle \frac{1}{3!}}+{\displaystyle \frac{1}{5!}}+\dots }=$

The sum of the series $\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+\dots \infty$ is

Sum of the series $\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+\frac{80}{81}+\dots$ to $n$ terms is