Question

等差数列前n项和Sn Sm=k Sk=m 求Sm+k

Answer 1

{an}是等差数列:Sm=ma1+1/2m(m-1)d=kSk=ka1+1/2k(k-1)d=mkma1+1/2km(m-1)d=k^2mka1+1/2km(k-1)d=m^2相减得:1/2kmd(m-k)=(k-m)(k+m)即kmd=-2(k+m)所以有 Sm+k=(m+k)a1+1/2(m+k)(m+k-1)d=ma1+ka1+1/2d(m^2+mk-m+mk+k^2...