1 + a^(2) + a^(4) + a^(6) + .........+ a^(n-2).Here "n" is an even integer.
2006-10-06 21:42:51 · 7 answers · asked by rajesh bhowmick 2 in Science & Mathematics ➔ Mathematics
Again this is a GP
a1 = 1, r = a², n = n - 2
Sum, S = a1(1 - a^n)/(1-n)
= (a^(n - 2) - 1)/(1 + n)
2006-10-06 21:55:38 · answer #1 · answered by Lance 2 · 2⤊ 0⤋
This is a geometric progression of the form 1+r+r^2+...+r^(k-1), with r=a^2 and k=n/2. The sum of such a progression is (r^k-1)/(r-1), so plug a^2 in for r and you've got the sum.
2006-10-06 21:46:32 · answer #2 · answered by James L 5 · 2⤊ 0⤋
its a geometric prog with a^2 the multiplictive factor, in the infinite form its only defiened for modulus(a) < 1 but can be analytically extened to the whole plane with two simple poles at +-1 by the function 1/(1-a^2) or maybe its two simple poles at +-i by the function 1/(1+a^2)... according to most painter and decoratoring football coaches.
2006-10-07 08:06:38 · answer #3 · answered by yellowdog 2 · 0⤊ 0⤋
Oh wow
2006-10-06 21:49:55 · answer #4 · answered by Anonymous · 0⤊ 1⤋
sorry, abhi tak ap nai padaaya...!!!
2006-10-06 22:41:07 · answer #5 · answered by Bluffmaster 3 · 1⤊ 0⤋
Well, "sum" might say it's this, but "sum" might say it's that.
2006-10-06 21:45:03 · answer #6 · answered by backinbowl 6 · 0⤊ 2⤋
sorry not defined..
2006-10-06 21:49:40 · answer #7 · answered by vaishak k 2 · 0⤊ 1⤋