$1+(1+3)+(1+3+5)+(1+3+5+7)+\ldots$ to $10$ terms $=$

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

If the sum of the first $15$ terms of the series ${\left( {\frac{3}{4}} \right)^3} + {\left( {1\frac{1}{2}} \right)^3} + {\left( {2\frac{1}{4}} \right)^3} + {3^3} + {\left( {3\frac{3}{4}} \right)^3} + \dots$ is equal to $225\,k$,then $k$ is equal to

If $\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^i {\sum\limits_{k = 1}^j {1 = 560} } } $,then the value of $n$ is:

Let $\{a_{n}\}_{n=0}^{\infty}$ be a sequence such that $a_{0}=a_{1}=0$ and $a_{n+2}=2a_{n+1}-a_{n}+1$ for all $n \geq 0$. Then,$\sum\limits_{n=2}^{\infty} \frac{a_{n}}{7^{n}}$ is equal to

The product $(32)(32)^{1/6}(32)^{1/36} \dots \infty$ is