How do you find the maclaurin series expansion of #f(x) = (e^x) + (e^(2x))#?

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Answer 1 · 2017-06-21T08:09:49

#e^x+e^(2x) = sum_(n=0)^oo (1+2^n)x^n/(n!) #

We have that:

#d/dx e^x = e^x#

so that clearly:

#d^n/dx^n e^x = e^x#

and:

#[d^n/dx^n e^x]_(x=0) = 1#

which means the MacLaurin series for #e^x# is:

#e^x = sum_(n=0)^oo x^n/(n!)#

and substituting #2x# for #x#:

#e^(2x) = sum_(n=0)^oo (2x)^n/(n!) = sum_(n=0)^oo 2^nx^n/(n!) #

Then:

#e^x+e^(2x) = sum_(n=0)^oo x^n/(n!) + sum_(n=0)^oo 2^nx^n/(n!) #

and grouping the terms of the same index:

#e^x+e^(2x) = sum_(n=0)^oo (1+2^n)x^n/(n!) #