已知数列{an}的通项公式an=2n+1⼀[n(n+1)]^2,求它的前n项和Sn为

2025-01-06 12:16:33
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回答1:

变通项公式
an=[(n+1)^2-n^2]/[n(n+1)]^2=1/n^2-1/(n+1)^2
a1=1-1/2^2
a2=1/2^2-1/3^2
a3=1/3^2-1/4^2

an=1/n^2-1/(n+1)^2
Sn=1-1/2^2+1/2^2-1/3^2+...+1/n^2-1/(n+1)^2=1-[1/(n+1)^2]