已知对于任意正整数n,都有a1+a2+…+an=n3,则1/a2−1+1/a3−1+…+1/a100−1=_.
问题描述:
已知对于任意正整数n,都有a1+a2+…+an=n3,则
+1
a2−1
+…+1
a3−1
=______. 1
a100−1
答
∵当n≥2时,有a1+a2+…+an-1+an=n3,a1+a2+…+an-1=(n-1)3,两式相减,得an=3n2-3n+1,∴1an−1=13n(n−1)=13(1n−1-1n),∴1a2−1+1a3−1+…+1a100−1,=13(1-12)+13(12-13)+…+13(199-1100),=13(1-11...