$2.5 + 5.9 + 8.13 + 11.17 + \ldots$ to $10$ terms $=$

The sum of $(1^2-1+1)(1!) + (2^2-2+1)(2!) + \ldots + (n^2-n+1)(n!)$ is

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 -

Let $ABC$ be an equilateral triangle with side length $a$. $A$ new triangle is formed by joining the midpoints of all sides of the triangle $ABC$,and the same process is repeated infinitely many times. If $P$ is the sum of perimeters and $Q$ is the sum of areas of all the triangles formed in this process,then:

If $\frac{2+4+6+8+\dots+\text{upto } n \text{ terms}}{1+3+5+7+\dots+\text{upto } n \text{ terms}} = \frac{37}{36}$,then $n = $

The sum of the series,$\frac{1}{2 \cdot 3} \cdot 2 + \frac{2}{3 \cdot 4} \cdot 2^{2} + \frac{3}{4 \cdot 5} \cdot 2^{3} + \ldots$ up to $n$ terms is