若an是等差数列,则Sn,S2n-Sn,S3n-S2n,…也成等差数列,求公差公差为dn^2 求证明过程 详细点
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若an是等差数列,则Sn,S2n-Sn,S3n-S2n,…也成等差数列,求公差公差为dn^2 求证明过程 详细点
若an是等差数列,则Sn,S2n-Sn,S3n-S2n,…也成等差数列,求公差
公差为dn^2 求证明过程 详细点
若an是等差数列,则Sn,S2n-Sn,S3n-S2n,…也成等差数列,求公差公差为dn^2 求证明过程 详细点
Sn=na1+n(n-1)d/2,
S2n=2na1+2n(2n-1)d/2,
S2n-Sn=na1+n(3n-1)d/2,
(S2n-Sn)-Sn=n²d,
k>1时,
[Skn -S(k-1)n]-[S(k-1)n -S(k-2)n]
={a[(k-1)n+1] +a[(k-1)n+2]+...+a[kn] } - {a[(k-2)n+1] +a[(k-2)n+2]+...+a[(k-1)n] }
={a[(k-1)n+1] -a[(k-2)n+1] }+ {a[(k-1)n+2] -a[(k-2)n+2]}+...+{a[kn] -a[(k-1)n] }
=nd+nd+...+nd 总共n项
=n²d,
所以从Sn开始就是等差.
Sn=na1+n(n-1)d/2,
S2n=2na1+2n(2n-1)d/2,
S2n-Sn=na1+n(3n-1)d/2,
(S2n-Sn)-Sn=n²d,
k>1时，
[Skn -S(k-1)n]-[S(k-1)n -S(k-2)n]
={a[(k-1)n+1] +a[(k-1)n+2]+...+a[kn] } - {a[(k-2)n+...
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Sn=na1+n(n-1)d/2,
S2n=2na1+2n(2n-1)d/2,
S2n-Sn=na1+n(3n-1)d/2,
(S2n-Sn)-Sn=n²d,
k>1时，
[Skn -S(k-1)n]-[S(k-1)n -S(k-2)n]
={a[(k-1)n+1] +a[(k-1)n+2]+...+a[kn] } - {a[(k-2)n+1] +a[(k-2)n+2]+...+a[(k-1)n] }
={a[(k-1)n+1] -a[(k-2)n+1] }+ {a[(k-1)n+2] -a[(k-2)n+2]}+...+{a[kn] -a[(k-1)n] }
=nd+nd+...+nd 总共n项
=n²d,
所以从Sn开始就是等差。
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