Question

求解一个幂级数的和函数

Answer 1

∑ (n: 1-> ∞ ) [(2n-1)/3^n ] x^(2n)
=∑ (n: 1-> ∞ ) (2n-1) (x/√3)^(2n)
=∑ (n: 1-> ∞ ) 2n.(x/√3)^(2n) -∑ (n: 1-> ∞ ) (x/√3)^(2n)
=6x^2 /(3-x^2)^2 -x^2/(3-x^2)
=(x^4 +3x^2)/ (3-x^2)^2
=x^2.(3+x^2) /(3-x^2)^2
consider
1/(1-u^2) = 1+u^2+u^4+....
u= x/√3
1/[ 1- (x^2/3) ] = 1+∑ (n: 1-> ∞ ) (x/√3)^(2n)
3/(3-x^2) =1+ ∑ (n: 1-> ∞ ) (x/√3)^(2n)
∑ (n: 1-> ∞ ) (x/√3)^(2n) = x^2/(3-x^2)
1/(1-u^2) = 1+u^2+u^4+....
两边求导
2u/(1-u^2)^2 = 2u+ 4u^3+....
两边乘以 u
2u^2/(1-u^2)^2 = 2u^2+ 4u^4+....
u=x/√3
[2x^2/3] /(1-x^2/3 )^2 = ∑ (n: 1-> ∞ ) 2n.(x/√3)^(2n)
∑ (n: 1-> ∞ ) 2n.(x/√3)^(2n) = 6x^2 /(3-x^2)^2