已知等比数列{an}的公比为q,前n项和为Sn,求[Sn*Sn+2-(Sn+1)^2]/[an*an+2]
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已知等比数列{an}的公比为q,前n项和为Sn,求[Sn*Sn+2-(Sn+1)^2]/[an*an+2]
已知数列{an}为等比数列,Sn为前n项之和.若S3S4-S4^2=-16,a2a4=32,求S4的值.
已知数列{an}为等比数列,Sn为前n项之和.若S3S4-S4^2=-16,a2a4=32,求S4的值.
![已知等比数列{an}的公比为q,前n项和为Sn,求[Sn*Sn+2-(Sn+1)^2]/[an*an+2]](/uploads/image/z/7631002-10-2.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%85%AC%E6%AF%94%E4%B8%BAq%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E6%B1%82%5BSn%2ASn%2B2-%EF%BC%88Sn%2B1%EF%BC%89%5E2%5D%2F%5Ban%2Aan%2B2%5D)
1)设an=a1*q^(n-1),则有
Sn=a1*(1-q^n)/(1-q),
[Sn*Sn+2-(Sn+1)^2]
=a1^2*{(1-q^n)*[1-q^(n+2)]-[1-q^(n+1)]^2}/(1-q)^2
=-a1^2*q^n,
an*an+2=a1^2*q^2n
所以[Sn*Sn+2-(Sn+1)^2]/[an*an+2] =-1/q^n
2)a2*a4=a3^2=32,所以a3=4√2,
得到 S4-S3=4√2,
由S3S4-S4^2=-16=S4(S3-S4)=-4√2S4,
得到S4=2√2.
注:本题如果想要先求的通项将会异常繁琐,技巧点在于利用中项性质求得a3,再变换为S4-S3
Sn=a1*(1-q^n)/(1-q),
[Sn*Sn+2-(Sn+1)^2]
=a1^2*{(1-q^n)*[1-q^(n+2)]-[1-q^(n+1)]^2}/(1-q)^2
=-a1^2*q^n,
an*an+2=a1^2*q^2n
所以[Sn*Sn+2-(Sn+1)^2]/[an*an+2] =-1/q^n
2)a2*a4=a3^2=32,所以a3=4√2,
得到 S4-S3=4√2,
由S3S4-S4^2=-16=S4(S3-S4)=-4√2S4,
得到S4=2√2.
注:本题如果想要先求的通项将会异常繁琐,技巧点在于利用中项性质求得a3,再变换为S4-S3
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