The  sum of the series $\frac{1}{1·2}-\frac{1}{2·3}+\frac{1}{3·4}-\mathrm{....}$upto $\infty$ is equal to

$\text{ Let }{S}_n=1·(n-1)+2·(n-2)+3·(n-3)+\dots +(n-1)·1, n⩾4.$ 

$\text{The sum }\sum _{\mathrm{n}=4}^{\infty }\left(\frac{2{ \mathrm{S}}_{\mathrm{n}}}{\mathrm{n}!}-\frac{1}{(\mathrm{n}-2)!}\right)\text{ is equal to :}$ 

On simplification the product

$\left({\mathrm{x}}_1-{\mathrm{y}}_1\right)\left({\mathrm{x}}_2+{\mathrm{y}}_2\right)...\left({\mathrm{x}}_{10} + {\mathrm{y}}_{10}\right)$

How many such terms are there which will have only single $\mathrm{x}$ and rest $\mathrm{y}$'s ?

Let $<a_n>$ be a sequence such that $a_1+a_2+...+a_n=\frac{n^2+3n}{\left(n+1\right)\left(n+2\right)}$. If $28\sum _{k=1}^{10}\frac{1}{a_k}=p_1 p_2 p_{3 }... p_m$, where $p_1, p_2, ... p_m$ are the first $m$ prime numbers, then $m$ is equal to