Find the first four nonzero terms of the Maclaurin series for the function by using an appropriate substitution in a known Maclaurin series and performing any algebraic operations that are required.

Question

Find the first four nonzero terms of the Maclaurin series for the function  by using an appropriate substitution in a known Maclaurin series and performing any algebraic operations that are required.

Quizplus · Accepted Answer

We can use the Maclaurin series for $e^x$ and make the substitution $x=3t^2-1$. Then we have:
$$e^{3t^2-1}=\sum_{n=0}^\infty\frac{(3t^2-1)^n}{n!}$$
To find the first four nonzero terms, we can plug in $n=0,1,2,3$ and simplify:
\begin{align*}
n&=0: \quad \frac{(3t^2-1)^0}{0!}&=1 \
n&=1: \quad \frac{(3t^2-1)^1}{1!}&=3t^2-1 \
n&=2: \quad \frac{(3t^2-1)^2}{2!}&=\frac{9}{2}t^4-3t^2+1 \
n&=3: \quad \frac{(3t^2-1)^3}{3!}&=\frac{81}{2}t^6-27t^4+\frac{27}{2}t^2-\frac{1}{2} \
\end{align*}
Now we replace $t$ with $x$ to get the terms in terms of the original variable, and we get:
$$1+(3x^2-1)+\frac{9}{2}x^4-3x^2+1+\frac{81}{2}x^6-27x^4+\frac{27}{2}x^2-\frac{1}{2}$$
Simplifying, we get:
$$\frac{81}{2}x^6-24x^4+\frac{27}{2}x^2$$
The first four nonzero terms are $\dfrac{81}{2}x^6, -24x^4, \dfrac{27}{2}x^2, 0$, so the answer is D.

Find the First Four Nonzero Terms of the Maclaurin Series