已知等比数列的前n项,前2n项,前3n项.求证Sn^2+S2n^2=Sn(S2n+S3n)
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已知等比数列的前n项,前2n项,前3n项.求证Sn^2+S2n^2=Sn(S2n+S3n)
已知等比数列的前n项,前2n项,前3n项.求证Sn^2+S2n^2=Sn(S2n+S3n)
已知等比数列的前n项,前2n项,前3n项.求证Sn^2+S2n^2=Sn(S2n+S3n)
证明：∵已知等比数列的前n项,前2n项,前3n项
∴S[n]=a[1](1-q^n)/(1-q)
S[2n]=a[1][1-q^(2n)]/(1-q)
S[3n]=a[1][1-q^(3n)]/(1-q)
∵S[n]^2+S[2n]^2
=[a[1](1-q^n)/(1-q)]^2+{a[1][1-q^(2n)]/(1-q)}^2
=a[1]^2{1-2q^n+q^(2n)+1-2q^(2n)+q^(4n)}/(1-q)^2
=a[1]^2{2-2q^n-q^(2n)+q^(4n)}/(1-q)^2
又∵S[n](S[2n]+S[3n])
=[a[1](1-q^n)/(1-q)]{a[1][1-q^(2n)]/(1-q)+a[1][1-q^(3n)]/(1-q)}
=a[1]^2{(1-q^n)[1-q^(2n)]+(1-q^n)[1-q^(3n)]}/(1-q)^2
=a[1]^2{1-q^n-q^(2n)+q^(3n)+1-q^n-q^(3n)+q^(4n)}/(1-q)^2
=a[1]^2{2-2q^n-q^(2n)+q^(4n)}/(1-q)^2
∴S[n]^2+S[2n]^2=S[n](S[2n]+S[3n])
证明：∵已知等比数列an的前n项，前2n项，前3n项
设首项伟a1,公比为q
∴S[n]=a[1](1-q^n)/(1-q)
S[2n]=a[1][1-q^(2n)]/(1-q)
S[3n]=a[1][1-q^(3n)]/(1-q)
∵S[n]^2+S[2n]^2
=[a[1](1-q^n)/(1-q)]^2+{a[1][1-q^...
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证明：∵已知等比数列an的前n项，前2n项，前3n项
设首项伟a1,公比为q
∴S[n]=a[1](1-q^n)/(1-q)
S[2n]=a[1][1-q^(2n)]/(1-q)
S[3n]=a[1][1-q^(3n)]/(1-q)
∵S[n]^2+S[2n]^2
=[a[1](1-q^n)/(1-q)]^2+{a[1][1-q^(2n)]/(1-q)}^2
=a[1]^2{1-2q^n+q^(2n)+1-2q^(2n)+q^(4n)}/(1-q)^2
=a[1]^2{2-2q^n-q^(2n)+q^(4n)}/(1-q)^2
又∵S[n](S[2n]+S[3n])
=[a[1](1-q^n)/(1-q)]{a[1][1-q^(2n)]/(1-q)+a[1][1-q^(3n)]/(1-q)}
=a[1]^2{(1-q^n)[1-q^(2n)]+(1-q^n)[1-q^(3n)]}/(1-q)^2
=a[1]^2{1-q^n-q^(2n)+q^(3n)+1-q^n-q^(3n)+q^(4n)}/(1-q)^2
=a[1]^2{2-2q^n-q^(2n)+q^(4n)}/(1-q)^2
∴S[n]^2+S[2n]^2=S[n](S[2n]+S[3n])
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