Question

数学:求∑(∞,n=1)n(n+1)/2^n 的和.

Answer 1

考虑幂级数∑(∞,n=1)n(n+1)x^n,
它的收敛区间为(-1,1),
设其和函数为S(x),
则S(x)=∑(∞,n=1)n(n+1)x^n
=x(∑(∞,n=1)x^(n+1))′
=x(x²/(1-x))″
=2x/(1-x)³.
故
∑(∞,n=1) n(n+1)/2^n
=S(1/2)
=(2*1/2)/(1-1/2)³
=8.

Answer 2

∑(∞,n=1)n(n+1)/2^n
consider
1/(1-x) = 1+x+x^2+....
1/(1-x)^2 = 1+2x+3x^2+... (两边取导)
x/(1-x)^2 = x + 2x^2+3x^3+....
x=1/2
∑(∞,n=1)n/2^n = 2
------------------------------
x/(1-x)^2 = x + 2x^2+3x^3+....
[(1-x)^2 + 2x(1-x) ]/(1-x)^4 = 1+ 2^2.x +3^2.x^2 +.... (两边取导)
(-x^2+1)/(1-x)^4 = 1+ 2^2.x +3^2.x^2 +....
x(-x^2+1)/(1-x)^4 = 1^2.x+ 2^2.x^2 +3^2.x^3 +....
x=1/2
∑(∞,n=1)n^2/2^n = (1/2) ( -1/4 +1)/( 1/2)^4 = 6
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∑(∞,n=1)n(n+1)/2^n
=∑(∞,n=1)n^2/2^n +∑(∞,n=1)n/2^n
=6+2
=8

Answer 3

1