Sequences and S...

Let $$S_{n}$$ denote the sum of cubes of the first $$n$$ natural numbers and $$s_{n}$$ denote the sum of the first $$n$$ natural numbers. Then $$\sum_{r=1}^{n}\dfrac{S_{r}}{S_{r}}$$ is equal to 

The value of $$\alpha$$ for which $$\sum_{n=1}^{\infty}A_n=\frac {8}{3}$$ is-

The value of $$\sum_{r=1}^{n}\left \{ \left ( 2r-1 \right )a+\dfrac{1}{b^{r}} \right \}$$ is equal to 

Find the sum of the products of every pair of the first $$n$$ natural numbers.

The numbers 1, 2, ..., 100 are arranged in the squares of an table in the following way: the numbers 1, ... , 10 are in the bottom row in increasing order,  numbers 11, ... ,20 are in the next row in increasing order, and so on. One can choose any number and two of its neighbors in two opposite directions (horizontal, vertical, or diagonal). Then either the number is increased by 2 and its neighbors are decreased by 1, or the number is decreased by 2 and its neighbors are increased by 1. After several such operations the table again contains all the numbers 1, 2, ... , 100. Prove that they are in the original order.

If n is an odd integer greater than or equal to 1 then the value of $$n^{3}-\left ( n-1 \right )^{3}+\left ( n-2 \right )^{3}-...+\left ( -1 \right )^{n-1}.1^{3}$$ is

Sum of the series $$11+23+45+87 ...$$ up to $$n$$ terms

$$\displaystyle 1^{3}-2^{3}+3^{3}-4^{3}+...+9^{3}$$ equals

The sum of all possible product of $$1^{st}$$ $$n$$ natural numbers taken two at a time is