$\begin{array}{1 1} x^4 \large\frac{1-x^{2n}}{1-x^2} \\ x^2 \large\frac{1-x^{n}}{1-x^2} \\x^2 \large\frac{1-x^{2n}}{1+x^{2n}} \\ x^2 \large\frac{1-x^{2n}}{1-x^2}\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- Sum of $n$ terms of a G.P.$=S_n=a.\large\frac{1-r^n}{1-r}$ where $a=$ first term and $r$=common ratio.

Given G.P.$=x^2,x^4,x^6.........$

In this G.P. first term$=x^2$ and common ratio$=\large\frac{x^4}{x^2}$$=x^2$

We know that the sum of $n$ terms of a G.P.$=a.\large\frac{1-r^n}{1-r}$

$\therefore\:S_n=x^2.\large\frac{1-(x^2)^n}{1-x^2}$$=x^2.\large\frac{1-x^{2n}}{1-x^2}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...