What will be the sum of this progression
1 + a^(2) + a^(4) + a^(6) + ........ + a^(n-2).Here "n" is an even integer.
2006-10-07 21:27:31 · 5 answers · asked by rajesh bhowmick 2 in Science & Mathematics ➔ Mathematics
Let Sum be S. Therefore
S=1 + a^(2) + a^(4) + a^(6) + ........ + a^(n-2) ...(1)
a^2S= a^(2) + a^(4) + a^(6) + ........ + a^(n-2) + a^n ...(2)
Substracting (2) from (1)
S(1-a^2)=1 - a^n
S=(1-a^n)/((1-a^2)
2006-10-07 22:46:07 · answer #1 · answered by LodhiRajput 3 · 0⤊ 0⤋
n=12
2006-10-08 04:36:56 · answer #2 · answered by Anonymous · 1⤊ 0⤋
s= a (r^n-1)/(r-1) ... (1)
from this formula
we have,
sum =(a^((n-2)/2+1)-1)/(a-1)
we get this by subs the following in (1)
a=1;
r=a;
n=(n-2)/2+1.
2006-10-08 07:57:03 · answer #3 · answered by know it all 1 · 0⤊ 0⤋
I already told you, it's (a^n - 1)/(a²-1). Please read the answers you have been given before spamming the board with your useless pleas for help.
2006-10-08 05:01:26 · answer #4 · answered by Pascal 7 · 0⤊ 0⤋
i wish i could help, but am freazing in my office......sorry dude...hehehe
2006-10-08 04:37:45 · answer #5 · answered by Gossai 3 · 0⤊ 0⤋