等差数列前n项和Sn Sm=k Sk=m 求Sm+k
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等差数列前n项和Sn Sm=k Sk=m 求Sm+k
{an}是等差数列:
Sm=ma1+1/2m(m-1)d=k
Sk=ka1+1/2k(k-1)d=m
kma1+1/2km(m-1)d=k^2
mka1+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-k)
=[ma1+1/2d(m^2-m)]+[ka1+1/2d(k^2-k)]+1/2d*2mk
=k+m+dmk
=k+m-2(k+m)
=-(k+m)
Sm=ma1+1/2m(m-1)d=k
Sk=ka1+1/2k(k-1)d=m
kma1+1/2km(m-1)d=k^2
mka1+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-k)
=[ma1+1/2d(m^2-m)]+[ka1+1/2d(k^2-k)]+1/2d*2mk
=k+m+dmk
=k+m-2(k+m)
=-(k+m)
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