Calc 2 question?:Find a function whose Maclaurin expansion is 1 + x3 + x6/2! + x9/3! + x12/4! + ...

so I thought that it would be using e^x but I can't get the correct answer

1 Answer
Mar 26, 2018

The given Maclaurin Expansion is that of #e^(x^3)#.

Explanation:

We have:

#1 + x^3 + x^6/(2!) + x^9/(3!) + x^12/(4!) + ... + x^(3n)/(n!)#

Now recall that the formula for #e^x# is

#1 + x+ x^2/(2!) + x^3/(3!) + x^4/(4!) + .... + x^n/(n!)#

If we raise each term in the expansion of #e^x# to power #3# we get the maclaurin expansion in the given question. Therefore, our maclaurin series is #e^(x^3)#.

Hopefully this helps!