等比数列{an}S的前n项和Sn=2^n-1,则a1^2+a2^2+a3^2+…+an^2

2025-06-22 12:39:26
推荐回答(1个)
回答1:

Sn=2^n -1,
S(n-1)=2^(n-1) -1,

an=Sn-S(n-1)
=2^n -1-[2^(n-1) -1]
=2^n-2^(n-1)
=2^(n-1)

an^2=2^(2n-2)=4^n/4,
a(n+1)^2=4^(n+1)/4,
a(n+1)^2/an^2=4
an^2是以a1^2=1为首项,4为公比的等比数列;

S=(1-4^n)/(1-4)
=(4^n-1)/3
=4^n/3-1/3