Tom K. answered 04/04/21
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It is D.
The Maclaurin series for e^x is ∑x^n/n!
Thus, the Maclaurin series for e^(x^2) is replacing x with x^2, ∑(x^2)^n/n! = ∑x^2n/n!
and the Maclaurin series for x^2e^(x^2) is ∑x^2x^2n/n! = ∑x^(2n+2)/n!