Question

求幂级数[∞∑ n=1] [(2n+1)x^(2n)] / n!] 的和函数

答案是：[(2x^2+1)e^(x^2)-1] (-∞,)∞

Answer 1

逐项积分得: ∑ x^(2n+1)/n!=x∑ x^2n/n! =x(e^(x^2)-1) x属于(-∞,∞) 求导得: 原级数=(e^(x^2)-1)+2x^2e^(x^2) =(1+2x^2)e^(x^2)-1, x属于(-∞,∞)

Answer 2

[(2n+1)x^(2n)] =[(2n)x^(2n)] +[x^(2n)] ,从这个角度看应该可以