如图 求这个幂级数的和函数

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∑{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).