Question

数列{an}中,a1+a2+a3+…+an=(2^n)-1,则a1^2+a2^2+a3^2+…+an^2等于

Answer 1

S(n) = 2^n-1,

a(n+1) = S(n+1) - S(n) = 2^(n+1) - 1 - 2^n+1 = 2^n,

a(1) = S(1) = 2-1=1.

a(n) = 2^(n-1), n = 1,2,...

[a(1)]^2 + [a(2)]^2 + ... + [a(n)]^2

= 1 + 2^2 + ... + [2^(n-1)]^2

= 1 + 4 + ... + 4^(n-1)

= [4^n - 1]/[4-1]

= [4^n - 1]/3.