Question

已知等差数列｛an｝的首项a1=1,公差d=2/3，且bn=[(-1)^(n-1)]an*a(n+1),求数列｛bn｝的前n项和Sn。 谢谢啦

Answer 1

an=(2/3)n+1/3
bn=[(-1)^(n-1)]an*a(n+1)=[(-1)^(n-1)]*[(2/3)n+1/3][(2/3)n+1]
Sn-S(n-1)=bn=[(-1)^(n-1)]*[(2/3)n+1/3][(2/3)n+1]
Tn=Sn/(-1)^n -->Tn+T(n-1)=-(4/9)n^2-(8/9)n-1/3
(Tn+(2/9)n^2+(2/3)n+7/18)=-(T(n-1)+(2/9)(n-1)^2+(2/3)(n-1)+7/18)
T1=S1/(-1)^n=b1/(-1)^1=[(-1)^0]a1a2/(-1)^0=-5/3
-->Tn+(2/9)n^2+(2/3)n+7/18是以-1为比，-5/3为首项的数列有
Tn+(2/9)n^2+(2/3)n+7/18=(-1)^n*(5/3)
Sn/(-1)^n=-(2/9)n^2-(2/3)n-7/18+(-1)^n*(5/3)
Sn=(-1)^(n+1)((2/9)n^2+(2/3)n+7/18)+5/3