1 + a^(2) + a^(4) + a^(6) +......?

Question

What will be the sum of this progression?

Answer 1

This is a GP in a^2
so sum = (1-(a^2(^n)))(1-a^2) = (1-a^(2n))/(1-a^2)

if there are n terms.

in case infinite terms |a| < 1
and limit is 1/(1-a^2)
else it is inifinite

Answer 2

The progression is in the form of an infinite Geometric Progression.

First term, a(1) = 1
Common ratio, r = a²

Sum of this infinite GP = a(1 - r^n)/(1 - r)
For a < 1, then
sum of the GP = a/ (1 - r) [a^n << 1 and can be ignored}
= 1/(1 - a²)
for a>1 , then
sum of GP = (1 - a^(2n))/(1 - r)
Here nth term is the last term of GP

Answer 3

An infinite series that can be written in the form

a+ar+ar^2+ar^3+... (forever)

is called a geometric series. The geometric series _diverges_ if |r|<1 and _converges_ if |r|<1. If it converges, it converges to a/(1-r).

Your series is geometric with a=1 and r=a^2 so if |a|<1, it converges to
1/(1-(a^2)2)=1/(1-a^4).

Answer 4

1 / 1-a^2

Answer 5

without last term how i can find the sum..??because we r very busy persons..