数列{an}的前项n的和为Sn,存在常数A、B、C,使得an+Sn=An^2+Bn+C对任意正整数都成立.若数列{an}为等差数列数列{an}的前项n的和为Sn,存在常数A、B、C,使得an+Sn=An^2+Bn+C对任意正整数n都成立.若数列{an}

Question

数列{an}的前项n的和为Sn,存在常数A、B、C,使得an+Sn=An^2+Bn+C对任意正整数都成立.若数列{an}为等差数列
数列{an}的前项n的和为Sn,存在常数A、B、C,使得an+Sn=An^2+Bn+C对任意正整数n都成立.若数列{an}为等差数列,求证：3A-B+C＝0

Answer 1

证明,据题意,{an}为等差数列,不妨假设它的首项为a1,公差为k.所以：
Sn=n*a1+k*n*（n-1）/2
an+Sn=a1+（n-1）k+n*a1+k*n*（n-1）/2
=a1+nk-k+na1+(k/2)n^2-kn/2
=(k/2)n^2+(a1+k/2)n+(a1+k)
据题意,存在常数ABC对所有n成立,则显然：
A=k/2
B=a1+k/2
C=a1-k
这样简单演算,得到
3A-B+C=0
证毕.