求数列{n/2^n}的和S 求数列{(2n+3)-X^n}的和Tn
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求数列{n/2^n}的和S 求数列{(2n+3)-X^n}的和Tn
s(n)=1/2 + 2/2^2 + 3/2^3 + ...+ (n-1)/2^(n-1) + n/2^n,
2s(n) = 1/1 + 2/2 + 3/2^2 + ...+ (n-1)/2^(n-2) + n/2^(n-1),
s(n)=2s(n)-s(n)=1/1 + 1/2 + 1/2^2 + ...+ 1/2^(n-1) - n/2^n
=[1-(1/2)^n]/(1-1/2) - n/2^n
=2[1-1/2^n] - n/2^n
= 2 - (n+2)/2^n
x=1时,t(n) = (2+3) - 1 + (2*2+3) - 1 + ...+ (2*n+3) - 1 = n(n+1) + 3n - n = n(n+1) + 2n = n(n+3),
x不为1时,
t(n) = (2+3)-x + (2*2+3)-x^2 + ...+ (2*n+3)-x^n = n(n+1)+3n - x[1+x+...+x^(n-1)]
=n(n+4) - x[1-x^n]/(1-x)
2s(n) = 1/1 + 2/2 + 3/2^2 + ...+ (n-1)/2^(n-2) + n/2^(n-1),
s(n)=2s(n)-s(n)=1/1 + 1/2 + 1/2^2 + ...+ 1/2^(n-1) - n/2^n
=[1-(1/2)^n]/(1-1/2) - n/2^n
=2[1-1/2^n] - n/2^n
= 2 - (n+2)/2^n
x=1时,t(n) = (2+3) - 1 + (2*2+3) - 1 + ...+ (2*n+3) - 1 = n(n+1) + 3n - n = n(n+1) + 2n = n(n+3),
x不为1时,
t(n) = (2+3)-x + (2*2+3)-x^2 + ...+ (2*n+3)-x^n = n(n+1)+3n - x[1+x+...+x^(n-1)]
=n(n+4) - x[1-x^n]/(1-x)
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