数列an,满足Sn=n^2+2n+1,设bn=an*2^n,求bn的前n项和Tn

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数列an,满足Sn=n^2+2n+1,设bn=an*2^n,求bn的前n项和Tn
数列an,满足Sn=n^2+2n+1,设bn=an*2^n,求bn的前n项和Tn

数列an,满足Sn=n^2+2n+1,设bn=an*2^n,求bn的前n项和Tn
由Sn=n²+2n+1易得
a1=4 (当n=1)
an=2n-1 (当n≥2)
所以
b1=8 (当n=1)
bn=(2n-1)*2^n
Tn=8+3*2^2+5*2^3+7*2^4+...+(2n-1)*2^n
2Tn=16+ 3*2^3+5*2^4+...+(2n-3)*2^n+(2n-1)*2^(n+1)
两式相减得
Tn=8-12-2(2^3+2^4+2^5+...+2^n)+(2n-1)*2^(n+1)
化简得
Tn=(2n-1)*2^(n+1)-2^(n+2)+12
等比数列和等差数列相乘的时候,可用错位相减法

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