好难的数列题！已知an=(n+1)/[(n^2+2n)^2],求Sn

显示全部楼层 · 2012-12-2 18:46:31

an=(n+1)/[(n^2+2n)^2]=(n+1)/[n(n+2)]^2=1/4*[1/n^2-1/(n+2)^2]当n为偶数时sn=1/4*(1/1^2-1/3^2)+1/4*(1/2^2-1/4^2)+.............+1/4*[1/n^2-1/(n+2)^2]=1/4*[1/1^2-1/3^2+1/3^2-1/5^2+..........1/(n-3)^2-1/(n-1)^2]+1/4*[1/2^2-1/4^2+1/4^2-1/6^2+........+1/n^2-1/(n+2)^2]=1/4*[1/1^2-1/(n-1)^2]+1/4*[1/2^2-1/(n+2)^2]=1/4*[1-1/(...

千问 · 2012-12-2 18:46:31

解答；裂项求和即可4(n+1)/[(n^2+2n)^2]=[(n+2)2-n2]/[(n+2)2*n2]=(n+2)2[(n+2)2*n2]-n2/[(n+2)2*n2]=1/n2-1/(n+2)2∴ (n+1)/[...

千问 · 2012-12-2 18:46:31

{(2n-1)/2^n}= 2n/2^n - 1/2^n 对于后一部分 1/2^n , 其前n项和为等比数列求和 S2 = 1/2 + 1/2^2 + 1/2^3 + …… 1/2^n = (1/2) * [1 - (1/2)^n]/(1 - 1/2) = 1 - 1/2^n 对于前一部分 2n/2^n S1 = 2*(1/2 + 2...

千问 · 2012-12-2 18:46:31

an-1=n/[(n-1)^2+2（n-1)]^2=n/[(n-1)^2 *(n+1)^2]=1/4[1/(n-1)^2-1/(n+1)^2]所以 an=1/4[1/n^2-1/(n+2)^2]an-1=1/4[1/(n-1)^2-1/(n+1)^2]an-2=1/4[1/(n-2)^2-1/n^2]an-3=1/4[1/(n...