在数列{an},设S1=a1+a2+.+an,S2=an+1 +an+2 +.+a2n .S3=a2n+1 +a2n+2
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在数列{an},设S1=a1+a2+.+an,S2=an+1 +an+2 +.+a2n .S3=a2n+1 +a2n+2 +.+a3n.若数列{an}是等差
证明S1,S2,S3,也是等差
证明S1,S2,S3,也是等差
证:
{an}是等差数列,设公差为d.
S1=a1+a2+...+an=na1+n(n-1)d/2
S2=a(n+1)+a(n+2)+...+a(2n)=na(n+1)+n(n-1)d/2=n(a1+nd)+n(n-1)d/2
S3=a(2n+1)+a(2n+2)+...+a(3n)=na(2n+1)+n(n-1)d/2=n(a1+2nd)+n(n-1)d/2
S1+S3=na1+n(n-1)d/2+n(a1+2nd)+n(n-1)d/2
=2na1+2n^2d+2[n(n-1)d/2]
=2n(a1+nd)+2[n(n-1)d/2]
=2[n(a1+nd)+n(n-1)d/2]
=2S2
S3-S2=S2-S1
S1,S2,S3也是等差数列.
{an}是等差数列,设公差为d.
S1=a1+a2+...+an=na1+n(n-1)d/2
S2=a(n+1)+a(n+2)+...+a(2n)=na(n+1)+n(n-1)d/2=n(a1+nd)+n(n-1)d/2
S3=a(2n+1)+a(2n+2)+...+a(3n)=na(2n+1)+n(n-1)d/2=n(a1+2nd)+n(n-1)d/2
S1+S3=na1+n(n-1)d/2+n(a1+2nd)+n(n-1)d/2
=2na1+2n^2d+2[n(n-1)d/2]
=2n(a1+nd)+2[n(n-1)d/2]
=2[n(a1+nd)+n(n-1)d/2]
=2S2
S3-S2=S2-S1
S1,S2,S3也是等差数列.
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