等差数列〔An〕中A1=3公差d=2 Sn为前n项和求1/S1+1/S2+…+1/Sn的值

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等差数列〔An〕中A1=3公差d=2 Sn为前n项和求1/S1+1/S2+…+1/Sn的值

Sk=(a1+an)*n/2=[2a1+(n-1)d]*n/2=(2+n)n 所以1/Sn=[(1/n)-1/(2+n)]/2 所以S=1/S1+1/S2+...+1/Sn =(1/2)[(1-1/3)+(1/2-1/4)+(1/3-1/5)+……+(1/n)-1/(2+n)] =(1/2)[1+1/2-1/(n+1)-1/(n+2)] =3/4-[(2n+3)/2(n+1)(n+2)]