设等比数列{q^n-1}(|q|>1)的前n项和为Sn,则Sn+2/Sn的极限是

题目

设等比数列{q^n-1}(|q|>1)的前n项和为Sn,则Sn+2/Sn的极限是
A.1/q² B.1/q四次方 C.q² D.q四次方

答案

Sn=(1-q^n)/(1-q)
所以sn+2=(1-q^(n+2))/(1-q)
所以sn+2/sn=(1-q^(n+2))/(1-q^n)=q^2
即c