如图 求这个幂级数的和函数
1个回答

∑{0 ≤ n} (2n+1)/n!·x^(2n)的一个原函数为∑{0 ≤ n} x^(2n+1)/n!.

∑{0 ≤ n} x^(2n+1)/n!= x·∑{0 ≤ n} x^(2n)/n!= x·∑{0 ≤ n} (x²)^n/n!.

由e^x = ∑{0 ≤ n} x^n/n!即得xe^(x²) = x·∑{0 ≤ n} (x²)^n/n!.

于是∑{0 ≤ n} (2n+1)/n!·x^(2n) = (xe^(x²))' = e^(x²)+2x²e^(x²) = (1+2x²)e^(x²).