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)=?](/uploads/image/z/19638298-10-8.jpg?t=%E9%A6%96%E9%A1%B9%E4%B8%BA3%E5%85%AC%E5%B7%AE%E4%B8%BA2%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%281%2FS1%29%2B%281%2FS2%29%2B.%2B%281%2FSn%29%3D%3F)
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)]
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)]
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