若等比数列{an}的前n项和Sn=3×2^n+a(a为常数),则a1^2+a2^2+a3^2+…+an^2=?
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若等比数列{an}的前n项和Sn=3×2^n+a(a为常数),则a1^2+a2^2+a3^2+…+an^2=?
![若等比数列{an}的前n项和Sn=3×2^n+a(a为常数),则a1^2+a2^2+a3^2+…+an^2=?](/uploads/image/z/19955250-18-0.jpg?t=%E8%8B%A5%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%3D3%C3%972%5En%2Ba%28a%E4%B8%BA%E5%B8%B8%E6%95%B0%29%2C%E5%88%99a1%5E2%2Ba2%5E2%2Ba3%5E2%2B%E2%80%A6%2Ban%5E2%3D%3F)
当n>=2时,an=Sn-S(n-1)=(3×2^n+a)-(3×2^(n-1)+a)=3*2^(n-1)
因为{an}是等比数列,所以当n=1时,a1=S1=6+a要适合an=3*2^(n-1),所以6+a=3*2^(1-1) 即 a=-3
令T=a1^2+a2^2+a3^2+…+an^2=3*2^0+3*2^1+3*2^2+.+3*2^(n-1)
=3*(1+2^1+2^2+.+2^(n-1))=3(2^n-1)
因为{an}是等比数列,所以当n=1时,a1=S1=6+a要适合an=3*2^(n-1),所以6+a=3*2^(1-1) 即 a=-3
令T=a1^2+a2^2+a3^2+…+an^2=3*2^0+3*2^1+3*2^2+.+3*2^(n-1)
=3*(1+2^1+2^2+.+2^(n-1))=3(2^n-1)
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