Question

无穷级数求和 ∑(n^2)/(2^n)

n属于1~无穷

Answer 1

∑(n^2)/(2^n)
=∑(n^2)
*
(1/2)^n
把1/2看成x
则得到幂级数
∑(n^2)
*
x^n
显然收敛域为x∈[-1,1)
=
x
*∑(n^2)
*
x^(n-1)
又因为
∑
n
*x^n
=x
*∑
n
*x^(n-1)=
x
*
(∑
(1-->+∞)
x^n)′=x^2/(1-x)
所以∑(n^2)
*
x^(n-1)=(∑
n
*x^n
)′=
[x^2/(1-x)]′=(2x-x^2)/(1-x)^2
所以∑(n^2)
*
x^n=
x
*
(2x-x^2)/(1-x)^2=(2x^2-x^3)/(1-x)^2
把x=1/2带入得到:
∑(n^2)/(2^n)=3/2