Question

一道求和函数的题 :∞∑(2n-1)x^2(n-1)/2^n的和函数

Answer 1

∑{1 ≤ n} (2n-1)x^(2n-2)/2^n

= (∑{1 ≤ n} x^(2n-1)/2^n)'

= (∑{0 ≤ n} x^(2n+1)/2^(n+1))'

= 1/2·(x·∑{0 ≤ n} x^(2n)/2^n)'

= 1/2·(x·∑{0 ≤ n} (x²/2)^n)'

= 1/2·(x/(1-x²/2))'

= 1/2·(1/(1-x²/2)+x·(-1)·(-x)/(1-x²/2)²)

= 1/2·(1+x²/2)/(1-x²/2)²

= (2+x²)/(2-x²)²