等差數(shù)列a1=3 d=2 Sn為前n項和 求Mn=1/S1+1/S2+1/S3+.+1/Sn
等差數(shù)列a1=3 d=2 Sn為前n項和 求Mn=1/S1+1/S2+1/S3+.+1/Sn
數(shù)學人氣：329 ℃時間：2020-05-20 04:19:22
優(yōu)質(zhì)解答
an=3+(n-1)*2=1+2n,sn=n*(3+1+2n)/2=n(n+2);1/sn=[1/n-1/(n+2)]/2;mn=[1-1/3+1/2-1/4+1/3-1/5+...+1/(n-1)-1/(n+1)+1/n-1/(n+2)]/2=[1+1/2-1/(n+1)-1/(n+2)]/2=(3n^2+7n+3)/[2(n+1)(n+2)]
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