The sum of the infinite series $1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\frac{22}{3^{5}}+\ldots$ is equal to

Find the sum to $n$ terms of the series $1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + \ldots$

$\sum\limits_{m = 1}^n {{m^2}}$ is equal to

$1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\ldots+\left(1^2+2^2+\ldots+n^2\right)=$

The sum of the series: $(2)^2 + 2(4)^2 + 3(6)^2 + \dots$ up to $10$ terms is

Let $\{a_{n}\}_{n=1}^{\infty}$ be a sequence such that $a_{1}=1, a_{2}=1$ and $a_{n+2}=2a_{n+1}+a_{n}$ for all $n \geq 1$. Then the value of $47 \sum_{n=1}^{\infty} \frac{a_{n}}{2^{3n}}$ is equal to $.....$