数列{an}的前n项和Sn与通项公式an满足关系式Sn=nan+2n2-2n(n∈N*),则a100-a10=( ) A.-90 B.-180 C.-360 D.-400
问题描述:
数列{an}的前n项和Sn与通项公式an满足关系式Sn=nan+2n2-2n(n∈N*),则a100-a10=( )
A. -90
B. -180
C. -360
D. -400
答
∵数列{an}的前n项和Sn与通项公式an满足关系式Sn=nan+2n2-2n(n∈N*),∴Sn+1=(n+1)an+1+2(n+1)2-2(n+1)=(n+1)•an+1+2n2+2n,两式相减作差,得:an+1=Sn+1-Sn=(n+1)an+1+4n-nan,整理得:an+1=an-4,即...