The sum to infinite terms of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ is

Consider a sequence whose sum of first $n$ terms is given by $S_n = 4n^2 + 6n$,where $n \in N$. Then,the $15^{th}$ term $(T_{15})$ of this sequence is:

If the sum of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots + 2 \cdot (n-1)^2 + n^2$ (when $n$ is odd) is to be determined,given that for even $n$,the sum is $\frac{n(n+1)^2}{2}$,find the sum when $n$ is odd.

$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ is equal to

If $1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots = \frac{a-2 \cos \theta}{5+b \cos \theta}$ for some $a, b \in R$,then $(a-b)^2=$

For an odd integer $n \ge 1$,the value of $n^3 - (n-1)^3 + (n-2)^3 - (n-3)^3 + \dots + (-1)^{n-1} 1^3$ is: