Question

高一数列求和数列｛an｝前n项和Sn=2^n-1,求a1^2+a2^2+…an^2

Answer 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.