Question

McLauren Series question inside

Answer 1

Find the 1000th term of
f(x)=(x^2+4)e ^{2x}\]
A general form for the formula would be aprreciative as well

Answer 2

\[\begin{align*}f(x)&=e^{2x}(x^2+4)\\
&=2^0e^{2x}\left(x^2+0x+\frac{8}{2}\right)\\
f'(x)&=2e^{2x}(x^2+4)+e^{2x}(2x)\\&=2e^{2x}(x^2+x+4)\\&=2^1e^{2x}\left(x^2+x+\frac{8}{2}\right)\\
f''(x)&=2^2e^{2x}(x^2+x+4)+2e^{2x}(2x+1)\\&=2^2e^{2x}\left(x^2+2x+\frac{9}{2}\right)\\
f'''(x)&=2^3e^{2x}\left(x^2+2x+\frac{9}{2}\right)+2^2e^{2x}(2x+2)\\&=2^3e^{2x}\left(x^2+3x+\frac{11}{2}\right)\\
f^{(4)}(x)&=2^4e^{2x}\left(x^2+3x+\frac{11}{2}\right)+2^3e^{2x}\left(2x+3\right)\\&=2^4e^{2x}\left(x^2+4x+\frac{14}{2}\right)\\
\vdots\\
f^{(n)}(x)&=2^ne^{2x}\left(x^2+nx+\frac{?}{2}\right)\end{align*}\]
I'm still not sure about the pattern for that numerator in each fraction. We have the seqeunce \(\{8,8,9,11,14,\ldots\}\). Looks recursive to me...

Anyway, if you can figure it out, the Maclaurin series for a function f is given by the sum
\[\sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(0)x^n\]