Pls help & give a to the point answer?

Question

I want to know the sum of this progression
1 + a^(2) + a^(4 + a^(6) + ......... + a^(n-2).Here "n" is an even integer.

The progression is
1 + a^(2) + a^(4) + ...... + a^(n-2)

Answer 1

Let sum be S. Therefore
S=1 + a^(2) + a^(4) + a^(6) + ......... + a^(n-2) ....(1)
multiplying by a^2
(a^2)S=a^(2) + a^(4) + a^(6) + ......... + a^(n-2) + a^n ...(2)
Substracting (2) from (1)
(1-a^2)S=1-a^n
S=(1-a^n)/(1-a^2)

Answer 2

It's a geometric series with A1=1, q=a^2

Answer 3

given:
s(n)=1+a^2+a^2+.......+a^(n-2)

we want to calculate the sum of this expression for n as any even Integer

We use a trick, because we rewrite the expression i opposit order, and add it to the expression:


......s(n)=1+ a^2 + a^4+.......+a^(n-2)
....+s(n)= a^(n-2)+ a^(n-4)......+a^n + a^0
-------------------------------------------------------
..2*s(n)=1+a^n + a^n+........+a^n +1

Now we have because n is an even integer

..2*s(n)=2+ (n/2)(a^n) <=>

s(n)=1+(n/4)(a^n) <=>

s(n)=1+(n*(a^n))/4
==============

Answer 4

Note that:
Let [k=1, (n-2)/2]∑a^(2k) = S
Then a²S=S-1+a^n
So (a²-1)S=a^n-1
Therefore S=(a^n - 1)/(a²-1)

Answer 5

google search.