Question

If n is even, then in the expansion of ${{\left( 1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+... \right)}^{2}}$ , the coefficient of ${{x}^{n}}$ is
A. $\dfrac{{{2}^{n}}}{n!}$
B. $\dfrac{{{2}^{n}}-2}{n!}$
C. $\dfrac{{{2}^{n-1}}-1}{n!}$
D. $\dfrac{{{2}^{n-1}}}{n!}$ .

Answer 1

Hint:Use Taylor series for expansion of ${{e}^{x}}$ and ${{e}^{-x}}$ .Add the series and square then to get the series given. Now fix ${{e}^{2x}}$ and ${{e}^{-2x}}$ and add them. Substitute their values back to the squared series and simplify it to get a coefficient of ${{x}^{n}}$ .

Complete step-by-step answer:
We have been given the expansion of an expression. We need to find the coefficient of ${{x}^{n}}$ from the given expression.
We have been given, ${{\left( 1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+... \right)}^{2}}$ .
According to Taylor series which is an expansion of some function into an infinite sum of terms, where each term has a larger exponent like $x,{{x}^{2}},{{x}^{3}}.....etc$ .
By using Taylor series, we can say that,
${{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+......\to (1)$ .
Similarly, ${{e}^{-x}}=1-\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}.....\to (2)$ .
Now let us add (1) and (2).
${{e}^{x}}+{{e}^{-x}}=\left( 1+\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+..... \right)+\left( 1-\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+.... \right)$ .
${{e}^{x}}+{{e}^{-x}}=2\left( 1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{6}}}{6!}+..... \right)$ .
Now let us find the square of $\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)$ .
${{\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)}^{2}}=\dfrac{1}{4}\left( {{e}^{2x}}+2{{e}^{x}}.{{e}^{-x}}+{{e}^{-2x}} \right)$ .
$\because {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+b$ .
${{\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)}^{2}}=\dfrac{1}{4}\left[ {{e}^{2x}}+{{e}^{-2x}}+2 \right]\to (4)$ .
We know ${{e}^{x}}$ and ${{e}^{-x}}$ from Taylor’s sense. Thus let us find the expansion of ${{e}^{2x}}$ and ${{e}^{-2x}}$ . Thus substitute $(2x)$ in place of $x$ .
$\begin{align}
  & {{e}^{2x}}=1+\dfrac{2x}{1!}+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+\dfrac{{{\left( 2x \right)}^{3}}}{3!}+.....\to (5) \\
 & {{e}^{-2x}}=1-\dfrac{2x}{1!}+\dfrac{{{\left( 2x \right)}^{2}}}{2!}-\dfrac{{{\left( 2x \right)}^{3}}}{3!}+.....\to (6) \\
\end{align}$
Bow let us add (5) and (6). We get,
\[\begin{align}
  & {{e}^{2x}}+{{e}^{-2x}}=\left( 1+\dfrac{2x}{1!}+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+\dfrac{{{\left( 2x \right)}^{3}}}{3!}+.... \right)+\left( 1-\dfrac{2x}{1!}+\dfrac{{{\left( 2x \right)}^{2}}}{2!}-\dfrac{{{\left( 2x \right)}^{3}}}{3!} \right) \\
 & {{e}^{2x}}+{{e}^{-2x}}=2\left( 1+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+\dfrac{{{\left( 2x \right)}^{4}}}{4!}+.... \right)\to (7) \\
\end{align}\]
Now let us substitute the value of (7) in (4)
$\begin{align}
  & =\dfrac{1}{4}\left( {{e}^{2x}}+{{e}^{-2x}}+2 \right)=\dfrac{1}{4}\left[ 2\left( 1+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+\dfrac{{{\left( 2x \right)}^{4}}}{4!}+.... \right)+2 \right] \\
 & =\dfrac{2}{4}\left[ 1+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+\dfrac{{{\left( 2x \right)}^{4}}}{4!}+....+1 \right] \\
 & =\dfrac{1}{2}\left[ 2+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+\dfrac{{{\left( 2x \right)}^{4}}}{4!}+.... \right]. \\
\end{align}$
Now, $2+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+\dfrac{{{\left( 2x \right)}^{4}}}{4!}+....$ is of the form $\left( \dfrac{{{2}^{n}}{{x}^{n}}}{n!} \right).$
Thus we can find,
$\dfrac{1}{2}\left[ 2+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+\dfrac{{{\left( 2x \right)}^{4}}}{4!}+..... \right]=\dfrac{1}{2}\times \left( \dfrac{{{2}^{n}}{{x}^{n}}}{n!} \right)$ .
$\dfrac{1}{2}.\dfrac{{{2}^{n}}}{n!}{{x}^{n}}=\dfrac{{{2}^{n}}\times {{2}^{-1}}}{n!}{{x}^{n}}$ .
$=\dfrac{{{2}^{n-1}}}{n!}{{x}^{n}}$ .
We were asked to find the coefficient of ${{x}^{n}}$ . We get,
$\dfrac{1}{2}\left[ 2+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+\dfrac{{{\left( 2x \right)}^{4}}}{4!}+.... \right]=\dfrac{{{2}^{n-1}}}{n!}{{x}^{n}}$ .
Hence the coefficient of ${{x}^{2}}=\dfrac{{{2}^{n-1}}}{n!}$ .
Thus we got the required value. Hence option (D) is the correct answer.

Note: We have series for the given expansion. You should be able to identify the given expansion. So it is important that you learn basic expansions of ${{e}^{x}},{{e}^{-x}},\cos x,{{\left( 1-x \right)}^{n}},{{\left( 1+x \right)}^{n}}etc$.Students should remember the expansion of ${{e}^{x}}=1+\dfrac{x}{1!}+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+......\to (1)$ and replacing x by -x we get expansion of ${{e}^{-x}}$.