已知数列an为首项a1≠0,公差为d≠0的等差数列,求Sn=1/a1a2+1/a2a3+……+1/ana(n+1)
人气:437 ℃ 时间:2020-05-11 12:42:42
解答
sn=1/d(1/a1-1/a2+1/a2-1/a3+.+1/an-1/a(n+1))
=1/d(1/a1-1/a(n+1))
=nd/da1a(n+1)
=n/a1a(n+1)
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