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

来源：百度知道 编辑：UC知道 时间：2024/05/30 16:18:20
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s(n) = a1 + a2 ...an = (2^n)-1
s(n-1) = =[2^(n-1)]-1
an = s(n) - s(n-1) = 2^(n-1)

a1 = 1
a2 = 2
a3 = 4
..

a1^2+a2^2+a3^2+…+an^2 = 1 + 4 + 4^2 .. .... + 4^(n-1) =
(4^n - 1)/3

a1+a2+a3+…+an=(2^n)-1
a1+a2+a3+…+an-1=2^(n-1)-1
两式相减
an=(2^n)-2^(n-1)=2^(n-1)
an^2=4^(n-1)
a1^2+a2^2+a3^2+…+an^2=
(4^n-1)/3

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.