Question

已知数列{an}的前n项和Sn=﹣n²+10n,求通项an=?

Answer 1

(1)由sn=-n^2+10n得,s(n-1)=-(n-1)^2+10(n-1),
两式相减,考虑到s(n)-s(n-1)=an得
an=-2n+11,a(n-1)=-2(n-1)+11,两式相减得,
an-a(n-1)=-2,故{an}是公差为-2的等差数列.
(2)a1=s1=-1^2+10=9,即a1=9,利用等差数列公式得
an=9-2(n-1),a2=7,a3=5,a4=3,a5=1,a6=-1,a7=-3,...
显然仅当n≥6时,an<0,
故当n≤6时,∣an∣=an,当n≥6时∣an∣=-an,即当n≤6时,｛∣an∣｝仍是公差为-2的等差数列,tn=9n-n(n-1),特别有t5=45-20=25.
当n≥6时,｛∣an∣｝是公差为2的等差数列.
tn=25+(n-5)*1+(n-5)(n-6),
即
当n≤6时,tn=10n-n^2,
当n≥6时,tn=25+(n-5)^2