设等差数列{an}的首项a1为a,公差d=2,前n项和为Sn(Ⅱ) 证明:n∈N*,Sn,Sn+1,Sn+2不构成等比数列.
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![设等差数列{an}的首项a1为a,公差d=2,前n项和为Sn(Ⅱ) 证明:n∈N*,Sn,Sn+1,Sn+2不构成等比数列.](/uploads/image/z/15190367-23-7.jpg?t=%E8%AE%BE%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%A6%96%E9%A1%B9a1%E4%B8%BAa%2C%E5%85%AC%E5%B7%AEd%EF%BC%9D2%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%28%E2%85%A1%29+%E8%AF%81%E6%98%8E%EF%BC%9An%E2%88%88N%2A%2CSn%2CSn%EF%BC%8B1%2CSn%EF%BC%8B2%E4%B8%8D%E6%9E%84%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%EF%BC%8E)
设等差数列{an}的首项a1为a,公差d=2,前n项和为Sn(Ⅱ) 证明:n∈N*,Sn,Sn+1,Sn+2不构成等比数列.
设等差数列{an}的首项a1为a,公差d=2,前n项和为Sn(Ⅱ) 证明:n∈N*,Sn,Sn+1,Sn+2不构成等比数列.
设等差数列{an}的首项a1为a,公差d=2,前n项和为Sn(Ⅱ) 证明:n∈N*,Sn,Sn+1,Sn+2不构成等比数列.
可以采用反证法.
等差数列的公式可以写成Sn=n[2a+2(n-1)],Sn+1=n(2a+2n),Sn+2=n[2a+2(n+1)]
若三者构成等比数列,则必有Sn*Sn+1=(Sn+2)^2,因此得到
[a+(n-1)][a+(n+1)]=(a+n)^2,得到a^2+2an+n^2-1=a^2+2an+n^2,因此推出-1=0,显然结论是错误的因此,Sn,Sn+1,Sn+2不构成等比数列.
这种题型一般都用反证法的.
Sn=(a1+an)n/2=(a+n-1)n
Sn+1=(a+n)(n+1)
Sn+2=(a+n+1)(n+2)
( Sn+1)*( Sn+1)=(a+n)(a+n)(n+1)(n+1)
Sn*Sn+2=(a+n-1)n(a+n+1)(n+2)
=[(a+n)(a+1...
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Sn=(a1+an)n/2=(a+n-1)n
Sn+1=(a+n)(n+1)
Sn+2=(a+n+1)(n+2)
( Sn+1)*( Sn+1)=(a+n)(a+n)(n+1)(n+1)
Sn*Sn+2=(a+n-1)n(a+n+1)(n+2)
=[(a+n)(a+1)-1]=[(n+1)(n+1)-1]
所以Sn的平方不可能等于Sn+1*Sn+2;
故存在任意正整数n,Sn,Sn+1,Sn+2不构成等比
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S²n+1=(n+1)²(n+a)² Sn×Sn+2=n(n+2)(n+a-1)(n+a+1) 设三者构成等比数列,S²n+1=Sn×Sn+2 即(n+1)²(n+a)²=n(n+2)(n+a-1)(n+a+1) (n+1)²(n+a)²=n(n+2)[(n+a)²-1] (n+1)²(n+a)²=n(n+2)(n+a)²-n(n+2) 移向得,(n+a)²[(n+1)²-n(n+2)]=-n(n+2) (n+a)²=-n(n+2) ∵n∈N*, ∴n>0 ∵(n+a)²≥0而-n(n+2)<0 两者矛盾,∴Sn,Sn+1,Sn+2不构成等比数列.