Question

设数列｛an｝的前N项和为Sn，已知1/Sn+1/S2+1/S3+......+1/Sn=n/(n+1)，求Sn

设数列｛an｝的前N项和为Sn，已知1/Sn+1/S2+1/S3+......+1/Sn=n/(n+1)，求Sn

Answer 1

解：由1/S1+1/S2+1/S3+......+1/Sn=n/(n+1)，知，当n=1时，s1=2,
当n≥2时1/S1+1/S2+1/S3+......+1/Sn-1=（n-1）/n，两式相减得，1/sn =1/[n(n+1)]
所以sn=n(n+1).,此式也适合s1.

Answer 2

1/Sn = (1/S1+1/S2+1/S3+......+1/Sn ) - (1/Sn+1/S2+1/S3+......+1/Sn-1) = n/(n+1) - (n-1)/n
即 1/Sn= 1/n(n+1)

所以Sn = n(n+1)
如 S1=2 , S2 = 6

呵呵，解数学题有时只要换个思路就好了