等差数列{an}前n项和为Sn,m≠n,Sn=m²,Sm=n²,求S(m+n)

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等差数列{an}前n项和为Sn,m≠n,Sn=m²,Sm=n²,求S(m+n)
等差数列{an}前n项和为Sn,m≠n,Sn=m²,Sm=n²,求S(m+n)

等差数列{an}前n项和为Sn,m≠n,Sn=m²,Sm=n²,求S(m+n)
Sn=m² = a1*n + d/2 *n(n-1)
Sm=n² = a1*m + d/2 *m(m-1)
Sn - Sm = m² - n² = a1(n-m) + d/2*(n-m)(n+m-1)
-(m+n) = a1 + d/2* (n+m-1)
S (m+n) = a1*(m+n) + d/2 *(m+n)(m+n-1) =-(m+n)²

等差数列{an}前n项和为Sn,m≠n,Sn=m²,Sm=n²,则有
Sn=na1+n(n-1)d/2=m²
m²/n=a1+(n-1)d/2
Sm=ma1+m(m-1)d/2=n²
n²/m=a1+(m-1)d/2
得m²/n-n²/m=(n-m)d/2
m²...

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等差数列{an}前n项和为Sn,m≠n,Sn=m²,Sm=n²,则有
Sn=na1+n(n-1)d/2=m²
m²/n=a1+(n-1)d/2
Sm=ma1+m(m-1)d/2=n²
n²/m=a1+(m-1)d/2
得m²/n-n²/m=(n-m)d/2
m²+mn+n²=-mnd/2
即mnd=-2(m²+mn+n²)
所以
S(m+n)
=(m+n)a1+(m+n)(m+n-1)d/2
=[ma1+m(m-1)d/2]+[na1+n(n-1)d/2]+mnd
=Sm+Sn+mnd
=m²+n²-2(m²+mn+n²)
=-m²-n²-2mn
=-(m+n)²

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