Answer :
We want to find the first four nonzero terms of
[tex]\[
f(x)=e^{6x}\sin x.
\][/tex]
To do this, we first write the Maclaurin (Taylor series at 0) expansions for [tex]\(e^{6x}\)[/tex] and [tex]\(\sin x\)[/tex].
The Maclaurin series for the exponential function is
[tex]\[
e^{6x}=\sum_{n=0}^{\infty}\frac{(6x)^n}{n!}=1+6x+18x^2+36x^3+\cdots,
\][/tex]
and for the sine function it is
[tex]\[
\sin x=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots.
\][/tex]
Because [tex]\(f(x)=e^{6x}\sin x\)[/tex], we multiply these two series term‐by-term. In general, the product is
[tex]\[
f(x)=\sum_{r\ge1} c_r x^r,
\][/tex]
with coefficients given by combining terms where the total power is [tex]\(r\)[/tex]. That is, a term from [tex]\(e^{6x}\)[/tex] of degree [tex]\(n\)[/tex] and a term from [tex]\(\sin x\)[/tex] of degree [tex]\(2k+1\)[/tex] contribute to the coefficient of [tex]\(x^{n+2k+1}\)[/tex].
Let’s find the first four nonzero terms:
1. [tex]\(\mathbf{x^1}\)[/tex]:
To obtain [tex]\(x^1\)[/tex], we need
[tex]\[
n+2k+1=1 \quad\Longrightarrow\quad n+2k=0.
\][/tex]
The only solution is [tex]\(n=0\)[/tex] and [tex]\(k=0\)[/tex]. The contribution is
[tex]\[
\frac{6^0}{0!}\cdot \frac{1}{1!} = 1\cdot 1=1.
\][/tex]
So, the [tex]\(x^1\)[/tex]-term is:
[tex]\[
x.
\][/tex]
2. [tex]\(\mathbf{x^2}\)[/tex]:
Now, for the [tex]\(x^2\)[/tex] term, we need
[tex]\[
n+2k+1=2 \quad\Longrightarrow\quad n+2k=1.
\][/tex]
The only possibility is [tex]\(n=1\)[/tex] and [tex]\(k=0\)[/tex]. Its contribution is
[tex]\[
\frac{6^1}{1!}\cdot \frac{1}{1!} = 6.
\][/tex]
Thus, the [tex]\(x^2\)[/tex]-term is:
[tex]\[
6x^2.
\][/tex]
3. [tex]\(\mathbf{x^3}\)[/tex]:
For the [tex]\(x^3\)[/tex] term, we have
[tex]\[
n+2k+1=3 \quad\Longrightarrow\quad n+2k=2.
\][/tex]
There are two possibilities:
- When [tex]\(k=0\)[/tex] then [tex]\(n=2\)[/tex]. The contribution is
[tex]\[
\frac{6^2}{2!}\cdot \frac{1}{1!} = \frac{36}{2} = 18.
\][/tex]
- When [tex]\(k=1\)[/tex] then [tex]\(n=0\)[/tex]. The contribution is
[tex]\[
\frac{6^0}{0!}\cdot \left(-\frac{1}{3!}\right)= 1\cdot\left(-\frac{1}{6}\right)=-\frac{1}{6}.
\][/tex]
Adding these contributions gives
[tex]\[
18-\frac{1}{6}=\frac{108}{6}-\frac{1}{6}=\frac{107}{6}.
\][/tex]
So, the [tex]\(x^3\)[/tex]-term is:
[tex]\[
\frac{107}{6}x^3.
\][/tex]
4. [tex]\(\mathbf{x^4}\)[/tex]:
For the [tex]\(x^4\)[/tex] term, we require
[tex]\[
n+2k+1=4 \quad\Longrightarrow\quad n+2k=3.
\][/tex]
The possibilities are:
- When [tex]\(k=0\)[/tex] then [tex]\(n=3\)[/tex]. The contribution is
[tex]\[
\frac{6^3}{3!}\cdot \frac{1}{1!} = \frac{216}{6} = 36.
\][/tex]
- When [tex]\(k=1\)[/tex] then [tex]\(n=1\)[/tex]. The contribution is
[tex]\[
\frac{6^1}{1!}\cdot \left(-\frac{1}{3!}\right)=6\cdot\left(-\frac{1}{6}\right)=-1.
\][/tex]
Thus, the [tex]\(x^4\)[/tex]-term has coefficient
[tex]\[
36-1=35.
\][/tex]
So, the [tex]\(x^4\)[/tex]-term is:
[tex]\[
35x^4.
\][/tex]
Collecting the terms, the Maclaurin series for [tex]\(f(x)\)[/tex] up to the first four nonzero terms is:
[tex]\[
f(x)=x+6x^2+\frac{107}{6}x^3+35x^4+\cdots.
\][/tex]
Among the options provided, the correct answer is:
B. [tex]\(x+6x^2+\frac{107}{6}x^3+35x^4+\cdots\)[/tex].
[tex]\[
f(x)=e^{6x}\sin x.
\][/tex]
To do this, we first write the Maclaurin (Taylor series at 0) expansions for [tex]\(e^{6x}\)[/tex] and [tex]\(\sin x\)[/tex].
The Maclaurin series for the exponential function is
[tex]\[
e^{6x}=\sum_{n=0}^{\infty}\frac{(6x)^n}{n!}=1+6x+18x^2+36x^3+\cdots,
\][/tex]
and for the sine function it is
[tex]\[
\sin x=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots.
\][/tex]
Because [tex]\(f(x)=e^{6x}\sin x\)[/tex], we multiply these two series term‐by-term. In general, the product is
[tex]\[
f(x)=\sum_{r\ge1} c_r x^r,
\][/tex]
with coefficients given by combining terms where the total power is [tex]\(r\)[/tex]. That is, a term from [tex]\(e^{6x}\)[/tex] of degree [tex]\(n\)[/tex] and a term from [tex]\(\sin x\)[/tex] of degree [tex]\(2k+1\)[/tex] contribute to the coefficient of [tex]\(x^{n+2k+1}\)[/tex].
Let’s find the first four nonzero terms:
1. [tex]\(\mathbf{x^1}\)[/tex]:
To obtain [tex]\(x^1\)[/tex], we need
[tex]\[
n+2k+1=1 \quad\Longrightarrow\quad n+2k=0.
\][/tex]
The only solution is [tex]\(n=0\)[/tex] and [tex]\(k=0\)[/tex]. The contribution is
[tex]\[
\frac{6^0}{0!}\cdot \frac{1}{1!} = 1\cdot 1=1.
\][/tex]
So, the [tex]\(x^1\)[/tex]-term is:
[tex]\[
x.
\][/tex]
2. [tex]\(\mathbf{x^2}\)[/tex]:
Now, for the [tex]\(x^2\)[/tex] term, we need
[tex]\[
n+2k+1=2 \quad\Longrightarrow\quad n+2k=1.
\][/tex]
The only possibility is [tex]\(n=1\)[/tex] and [tex]\(k=0\)[/tex]. Its contribution is
[tex]\[
\frac{6^1}{1!}\cdot \frac{1}{1!} = 6.
\][/tex]
Thus, the [tex]\(x^2\)[/tex]-term is:
[tex]\[
6x^2.
\][/tex]
3. [tex]\(\mathbf{x^3}\)[/tex]:
For the [tex]\(x^3\)[/tex] term, we have
[tex]\[
n+2k+1=3 \quad\Longrightarrow\quad n+2k=2.
\][/tex]
There are two possibilities:
- When [tex]\(k=0\)[/tex] then [tex]\(n=2\)[/tex]. The contribution is
[tex]\[
\frac{6^2}{2!}\cdot \frac{1}{1!} = \frac{36}{2} = 18.
\][/tex]
- When [tex]\(k=1\)[/tex] then [tex]\(n=0\)[/tex]. The contribution is
[tex]\[
\frac{6^0}{0!}\cdot \left(-\frac{1}{3!}\right)= 1\cdot\left(-\frac{1}{6}\right)=-\frac{1}{6}.
\][/tex]
Adding these contributions gives
[tex]\[
18-\frac{1}{6}=\frac{108}{6}-\frac{1}{6}=\frac{107}{6}.
\][/tex]
So, the [tex]\(x^3\)[/tex]-term is:
[tex]\[
\frac{107}{6}x^3.
\][/tex]
4. [tex]\(\mathbf{x^4}\)[/tex]:
For the [tex]\(x^4\)[/tex] term, we require
[tex]\[
n+2k+1=4 \quad\Longrightarrow\quad n+2k=3.
\][/tex]
The possibilities are:
- When [tex]\(k=0\)[/tex] then [tex]\(n=3\)[/tex]. The contribution is
[tex]\[
\frac{6^3}{3!}\cdot \frac{1}{1!} = \frac{216}{6} = 36.
\][/tex]
- When [tex]\(k=1\)[/tex] then [tex]\(n=1\)[/tex]. The contribution is
[tex]\[
\frac{6^1}{1!}\cdot \left(-\frac{1}{3!}\right)=6\cdot\left(-\frac{1}{6}\right)=-1.
\][/tex]
Thus, the [tex]\(x^4\)[/tex]-term has coefficient
[tex]\[
36-1=35.
\][/tex]
So, the [tex]\(x^4\)[/tex]-term is:
[tex]\[
35x^4.
\][/tex]
Collecting the terms, the Maclaurin series for [tex]\(f(x)\)[/tex] up to the first four nonzero terms is:
[tex]\[
f(x)=x+6x^2+\frac{107}{6}x^3+35x^4+\cdots.
\][/tex]
Among the options provided, the correct answer is:
B. [tex]\(x+6x^2+\frac{107}{6}x^3+35x^4+\cdots\)[/tex].
