Question

无穷级数求和 ∑(n^2)/(2^n)

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