The first term of a sequence is $2014$ . Each succeeding term is the sum of the cubes of the digits of the previous term. Then the ${2014}^{\mathrm{t}\mathrm{h}}$ term of the sequence  is

What should come at the place of question mark?

$46080,3840,384,48,8,2, ?$ 

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 ?

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

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