Question

In the expansion of \[{{\left( 1+x \right)}^{n}}\], what is the sum of even binomial coefficients?
(a)\[{{2}^{n}}\]
(b)\[{{2}^{n-1}}\]
(c)\[{{2}^{n+1}}\]
(d)None of these

Answer 1

Hint: Expand \[{{\left( 1+x \right)}^{n}}\] and mark it as equation (1). Now, put x = -x in (1) and simplify it. Add both the equations and put, x = 1, thus we will get the sum of even or even binomial coefficients.

Complete step-by-step answer:
We have been given the expression \[{{\left( 1+x \right)}^{n}}\]. The binomial expansion of \[{{\left( 1+x \right)}^{n}}\] is equal to, \[{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{r}}{{x}^{r}}\].
 \[{{\left( 1+x \right)}^{n}}=1+nx+\dfrac{n\left( n-1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n-1 \right)\left( n-2 \right)}{3!}{{x}^{3}}+......+\dfrac{n\left( n-1 \right)\left( n-2 \right)...\left( n-r+1 \right)}{r!}{{x}^{r}}+.....\]
We have been asked to find the sum of even binomial coefficient,
\[{{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+....+{}^{n}{{C}_{r}}{{x}^{r}}-(1)\]
Thus the even values are \[{{C}_{0}},{{C}_{2}},{{C}_{4}}....\] etc.
i.e. \[{{C}_{0}}+{{C}_{2}}+{{C}_{4}}+....\] is the sum of the even coefficients.
Thus, put x = -x in equation (1).
\[\begin{align}
  & {{\left( 1+\left( -x \right) \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}\left( -x \right)+{}^{n}{{C}_{2}}{{\left( -x \right)}^{2}}+.....+{}^{n}{{C}_{r}}{{\left( -x \right)}^{r}} \\
 & {{\left( 1-x \right)}^{n}}={}^{n}{{C}_{0}}-{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+\left( -1 \right){}^{n}{{C}_{r}}{{x}^{r}}-(2) \\
\end{align}\]
Now let us add equation (1) and (2).
\[{{\left( 1+x \right)}^{n}}+{{\left( 1-x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x-{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{2}}{{x}^{2}}+.....+{}^{n}{{C}_{r}}{{x}^{r}}-{}^{n}{{C}_{r}}{{x}^{r}}\]
Thus the above expression, after cancelling like terms becomes,
\[{{\left( 1+x \right)}^{n}}+{{\left( 1-x \right)}^{n}}=2{}^{n}{{C}_{0}}+2{}^{n}{{C}_{0}}{{x}^{2}}+2{}^{n}{{C}_{4}}{{x}^{4}}+.....\]
\[{{\left( 1+x \right)}^{n}}+{{\left( 1-x \right)}^{n}}=2\left[ {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{4}}{{x}^{4}}+..... \right]\]
Now let us put, x = 1.
\[\therefore {{\left( 1+x \right)}^{n}}+{{\left( 1-x \right)}^{n}}=2\left[ {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{4}}{{x}^{4}}+..... \right]\]
\[\begin{align}
  & {{\left( 1+1 \right)}^{n}}+{{\left( 1-1 \right)}^{n}}=2\left[ {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}{{\left( 1 \right)}^{2}}+{}^{n}{{C}_{4}}\times {{\left( 1 \right)}^{4}}+..... \right] \\
 & \therefore {{2}^{n}}+0=2\left[ {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{4}}+..... \right] \\
\end{align}\]
\[{{2}^{n}}=2\left[ {{C}_{0}}+{{C}_{2}}+{{C}_{4}}+.... \right]\left\{ \because \dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}} \right.\]
\[\dfrac{{{2}^{n}}}{2}={{C}_{0}}+{{C}_{2}}+{{C}_{4}}+.....=\dfrac{{{2}^{n}}}{2!}\]
\[\therefore \] The sum of even binomial coefficients = \[{{2}^{n-1}}\].
Thus we got the required answer.
\[\therefore \] Option (b) is the correct answer.

Note: This one is a very simple illustration of how we put some value of x and get the solution of the problem. It is very important how judiciously we exploit this property of the binomial expansion.