Let for $n =1,2, \ldots \ldots, 50, S _{ a }$ be the sum of the infinite geometric progression whose first term is $n ^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to

Let $S_1$ be the sum of areas of the squares whose sides are parallel to coordinate axes. Let $S_2$ be the sum of areas of the slanted squares as shown in the figure. Then, $\frac{S_1}{S_2}$ is equal to

Let ${a_n}$ be the ${n^{th}}$ term of the G.P. of positive numbers. Let $\sum\limits_{n = 1}^{100} {{a_{2n}}} = \alpha $ and $\sum\limits_{n = 1}^{100} {{a_{2n - 1}}} = \beta $, such that $\alpha \ne \beta $,then the common ratio is

The sum to infinity of the following series $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{2^3}}} + \frac{1}{{{3^3}}} + ........$, will be

In a geometric progression, if the ratio of the sum of first $5$ terms to the sum of their reciprocals is $49$, and the sum of the first and the third term is $35$ . Then the first term of this geometric progression is

The number of natural number $n$ in the interval $[1005, 2010]$ for which the polynomial. $1+x+x^2+x^3+\ldots+x^{n-1}$ divides the polynomial $1+x^2+x^4+x^6+\ldots+x^{2010}$ is