Question

已知各项均为正数的数列{an}满足:a1=1,an+1²-an²=2(n∈N*)

求数列{a∧n²/2∧n}的前n项和Sn

Answer 1

a1=1 a1²=1
a(n+1)²-an²=2
数列{an²}是以1为首项，2为公差的等差数列
an²=1+2(n-1)=2n-1
an²/2ⁿ=(2n-1)/2ⁿ
Sn=1/2 +3/2²+5/2³+...+(2n-1)/2ⁿ
(1/2)Sn=1/2²+3/2³+...+(2n-3)/2ⁿ+(2n-1)/2^(n+1)
Sn-(1/2)Sn=(1/2)Sn=1/2+1/2²+...+1/2ⁿ -(2n-1)/2^(n+1)
Sn=1+1/2+...+1/2^(n-1) -(2n-1)/2ⁿ
=1×(1-1/2ⁿ)/(1-1/2) -(2n-1)/2ⁿ
= 2- (2n+1)/2ⁿ