Question

等差数列前n项和Sn=m,前m项和Sm=n,n不等于m,则前m+n项和为多少?

Answer 1

既然是等差数列

Sn=na1+n(n-1)d/2=m

Sm=ma1+m(m-1)d/2=n

上式相减得

(n-m)a1+(n^2-n-m^2+m)d/2=m-n

a1+(n+m-1)d/2=-1

S（n+m)

=(n+m)a1+(n+m)(n+m-1)d/2

=(n+m)[a1+(n+m-1)d/2]

=-n-m