等差数列前n项和为Sn ,若Sn=Sm(m>
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等差数列前n项和为Sn ,若Sn=Sm(m>
等差数列前n项和为Sn ,若Sn=Sm(m>
等差数列前n项和为Sn ,若Sn=Sm(m>
Sn=na1+(1/2)n(n+1)d
Sm=ma1+(1/2)m(m+1)d
两式相减,得:
0=(n-m)a1+(1/2)d[(n²-m²)+(n-m)] 两边除以n-m,得:
a1+(1/2)(m+n+1)d=0
而:S(m+n)=(m+n)a1+(1/2)(m+n)(m+n+1)d
=(m+n)[a1+(1/2)(m+n+1)d]
=(m+n)×0
=0
美哟
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