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

The sum of the infinite series $(\frac{1}{3}+\frac{4}{7})+(\frac{1}{3^{2}}+\frac{1}{3}\times\frac{4}{7}+\frac{4^{2}}{7^{2}})+(\frac{1}{3^{3}}+\frac{1}{3^{2}}\times\frac{4}{7}+\frac{1}{3}\times\frac{4^{2}}{7^{2}}+\frac{4^{3}}{7^{3}}) + \dots$ is equal to -

The sum to infinity of the progression $9 - 3 + 1 - \frac{1}{3} + \dots$ is

Find the sum of the series $5^{2} + 6^{2} + 7^{2} + \ldots + 20^{2}$.

$\frac{10001 \times 100 !}{2 \times 1 !+5 \times 2 !+10 \times 3 !+\ldots+10001 \times 100 !}=$

If $\frac{1^3+2^3+3^3+\ldots \text{ upto } n \text{ terms}}{1 \cdot 3+2 \cdot 5+3 \cdot 7+\ldots \text{ upto } n \text{ terms}} = \frac{9}{5}$,then the value of $n$ is