首项为3公差为2的等差数列,(1/S1)+(1/S2)+.+(1/Sn)=?

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首项为3公差为2的等差数列,(1/S1)+(1/S2)+.+(1/Sn)=?
首项为3公差为2的等差数列,(1/S1)+(1/S2)+.+(1/Sn)=?

首项为3公差为2的等差数列,(1/S1)+(1/S2)+.+(1/Sn)=?
Sk=(a1+ak)*k/2=[2a1+(k-1)d]*k/2=(2+k)k
1/Sk=[(1/k)-1/(2+k)]/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)]

an = a1 + (n-1)d = 2n + 1
Sn = [n(a1 + an)]/2 = n(n+2)
1/Sn = 1/n(n+2) = [1/n - 1/(n+2)]/2
An = S1 + S2+ ... + Sn = 21/40 - (2n+3)/2(n+2)(n+1)

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