- 一数列题设正项数列{An}的前n项和为Sn,且存在正数t,使得对
- 设正项数列{An}的前n项和为Sn,且存在正数t,使得对所有的正整数n,t与An的等差中项和t与Sn的等比中项相等,求证数列{√Sn}为等差数列,并求{An}通项公式及前n项和
- (t +An)/2 = √(t*Sn)
==> t^2 +2t*An +An^2 = 4t*Sn ...(1)
n=1: S1=A1, ==> A1 = t
==> t^2 +2t*A(n-1) +A(n-1)^2 = 4t*S(n-1) ...(2)
(1)-(2):2t[An-A(n-1)]+An^2-A(n-1)^2=4t*[Sn-S(n-1)]=4t*An
==> [An+A(n-1),An-A(n-1)-2t] =0
==> An-A(n-1) = 2t
==> {An}为等差数列, A1=t, d =2t
An =2nt -t, Sn =t*n^2
题目应为:{An}为等差数列。