设Cn=(2n-1)*4^n-2,求数列{Cn}的前n项和Tn
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设Cn=(2n-1)*4^n-2,求数列{Cn}的前n项和Tn
设Cn=(2n-1)*4^n-2,求数列{Cn}的前n项和Tn
设Cn=(2n-1)*4^n-2,求数列{Cn}的前n项和Tn
见下图
c(n) = (2n-1)4^(n-2),
t(n) = c(1)+c(2)+c(3)+...+c(n-1)+c(n)
=(2*1-1)4^(-1) + (2*2-1)4^0 + (2*3-1)4^1 + ... + [2(n-1)-1]4^(n-3) + [2n-1]4^(n-2),
4t(n) = (2*1-1)4^0 + (2*2-1)4^1 + ... + [2(n...
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c(n) = (2n-1)4^(n-2),
t(n) = c(1)+c(2)+c(3)+...+c(n-1)+c(n)
=(2*1-1)4^(-1) + (2*2-1)4^0 + (2*3-1)4^1 + ... + [2(n-1)-1]4^(n-3) + [2n-1]4^(n-2),
4t(n) = (2*1-1)4^0 + (2*2-1)4^1 + ... + [2(n-1)-1]4^(n-2) + [2n-1]4^(n-1),
3t(n) = 4t(n) - t(n) = -(2*1-1)4^(-1) - 2*4^0 - 2*4^1 - ... - 2*4^(n-2) + [2n-1]4^(n-1)
= (2n-1)4^(n-1) + 1/4 - (2/4)[1+4+...+4^(n-1)]
= (2n-1)4^(n-1) + 1/4 - (1/2)[4^n - 1]/(4-1)
= (2n-1)4^(n-1) + 1/4 - [4^n - 1]/6
= [(12n-6-4)/6]4^(n-1) + 1/4 + 1/6
= [(6n-5)/3]4^(n-1) + 5/12,
t(n) = [(6n-5)/9]4^(n-1) + 5/36
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