xiangyun
花朵
共回答了16个问题采纳率:100% 举报
Sn=2/3n+1/2n^2
当n=1时,S1=2/3+1/2=a1
当n=2时,S2=2/6+1/8=a1+a2
当n=3时,S3=2/9+1/18=a1+a2+a3
...
当n=k时,Sk=2/3k+1/2k^2=a1+a2+a3+...ak
a1=S1
a2=S2-S1
a3=S3-S2
a4=S4-S3
...
ak=Sk-S(k-1)
=[2/3k+1/2k^2]-[2/3(k-1)+1/2(k-1)^2]
=2[1/3k-1/3(k-1)]+[1/2k^2-1/2(k-1)^2]
=-{2/3+(2k-1)/2[k(k-1)]}/[k(k-1)]
=[3-2k(2k+1)]/{6[k(k-1)]^2}
1年前
6