What is the Maclaurin Series for $f(x) = x^2ln(1+x^3)$ ?

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Answer 1 · 2017-05-01T22:04:48

$f(x) = x^5-x^8/2+x^11/3 - x^14/4...$

$" " = sum_(n=1)^(oo) \ (-1)^(n+1) \ x^(3n+2)/n$

We start with the well known Maclaurin Series for $ln(1+x)$ :

$ln (1+x) = x-x^2/2+x^3/3 - x^4/4 +...$

If we replace $x$ in the series by $x^3$ then we get:

$ln (1+x^3) = (x^3)-(x^3)^2/2+(x^3)^3/3 - (x^3)^4/4 + ...$
$" " = x^3-x^6/2+x^9/3 - x^12/4...$

So then we have:

$f(x) = x^2ln(1+x^3)$
$" " = x^2{x^3-x^6/2+x^9/3 - x^12/4... }$
$" " = x^5-x^8/2+x^11/3 - x^14/4...$
$" " = sum_(n=1)^(oo) \ (-1)^(n+1) \ x^(3n+2)/n$

What is the Maclaurin Series for f(x) = x^2ln(1+x^3) f(x) = x^2ln(1+x^3) ?