The sum of the series $\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{2^2}{x^4+1}+\ldots .+\frac{2^{100}}{x^{2^{100}}+1}$ when $x=2$ is

If a, b and c be three distinct real numbers in G.P. and a + b + c = xb, then x cannot be:

Let $a_1, a_2 \ldots \ldots a_{10}$ be an AP with common difference -3 and $b_1, b_2 \ldots \ldots, b_{10}$ be a GP with common ratio...

Let $a_1, a_2 \ldots . a_{30}$ be an A.P., $S=\sum_{i=1}^{30}$ ai and $T=\sum_{i=1}^{15} a_{(2 i-1)}$. If $a_5=27$ and $S-2 T=75$, then $a_{10}$ is equal to

Let $\mathrm{a}_1, \mathrm{a}_2 \ldots \ldots, \mathrm{a}_{21}$ be an AP such that $\sum_{\mathrm{n}=1}^{20} \frac{1}{\mathrm{a}_{\mathrm{n}} \mathrm{a}_{\mathrm{n}+1}}=\frac{4}{9}$. If the sum of this AP is 189 , then a6a16...

Let $S_n=1 \cdot(n-1)+2 \cdot(n-2)+3 \cdot(n-3)+\ldots \cdot+(n-1) \cdot n \geq 4$. The sum $\sum_{n=4}^{\infty}\left(\frac{2 S_n}{n!}-\frac{1}{(n-2)!}\right)$ is equal to :

If the value of $\left(1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\ldots . \text { upto } \infty\right)^{\log _{1008}\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^2}+\ldots \text { uptoss }\right)}$ is $l$, then $l^2$ is equal to $\_\_\_\_$ .

The sum $\sum_{k=1}^{20} k \frac{1}{2^k}$ is equal to :

Let $S_n$ be the sum of the first $n$ terms of ah arithmetic progression. If $S_{3 n}=3 S_{2 n}$, then the value of $...

If three distinct numbers $a, b, c$ are in G.P. and the equations $a x^2+2 b x+c=0$ and $d x^2+2 e x+f=0$ have $a$ common...