数列｛an｝的前n项和为sn 且a1=1/2 sn=n2/(n2-1)Sn-1+n/n+1（n≥2）

显示全部楼层 · 2013-5-9 16:57:22

S1=a1=1/2S2=(4/3)S1+(2/3)=4/3S3=(9/8)S2+(3/4)=9/4S4=(16/15)S2+(4/5)=16/5猜想: Sn=n2/(n+1)数学归纳法证明：假设：Sk=k2/(k+1)，k≥2，由已知，S(k+1) = {(k+1)2/[(k+1)2-1]}Sk + (k+1)/(k+2)= [(k+1)2/(k2+2k)][k2/(k+1)] + (k+1)/(k+2)= k(k+1)/(k+2) + (k+1)/(k+2)= (k+1)2/(k+2)= (k+1)2/[(k+...

千问 · 2013-5-9 16:57:22

Sn = [n^2/(n^2-1)]S(n-1) + n/(n+1)(n^2-1)Sn = n^2S(n-1) + n(n-1)[(n+1)/n]Sn - [n/(n-1)]S(n-1) = 1[(n+1)/n]Sn - (2/1)S1 = n-1[(n+1)/n]Sn = nSn= n^2/(n+1)S2 = 4/3S3=9...