Question

Find the value of \[{}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{4}}+\cdots +{}^{n}{{C}_{2\left[ {}^{n}/{}_{2} \right]}}\] , where $\left[ {} \right]$ denotes the greatest integer.
(a) ${{2}^{2n-1}}$
(b) ${{2}^{2n-1}}-1$
(c) ${{2}^{n-1}}$
(d) ${{2}^{n-1}}-1$

Answer 1

Hint: We will consider the binomial expansion of ${{(1+x)}^{n}}$ and ${{(1-x)}^{n}}$. The binomial expansion of ${{(1+x)}^{n}}$ and ${{(1-x)}^{n}}$ are as follows,
${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+\cdots +{}^{n}{{C}_{n}}{{x}^{n}}$
${{(1-x)}^{n}}={}^{n}{{C}_{0}}-{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}-\cdots +{{(-1)}^{n}}{}^{n}{{C}_{n}}{{x}^{n}}$
Then we will add these two expressions. We will substitute $x=1$ in the result obtained and then simplify the equation. Simplifying this equation will give us an expression where we are adding the even terms of a binomial expression.

Complete step-by-step answer:
We will first write the binomial expansion of ${{(1+x)}^{n}}$ and ${{(1-x)}^{n}}$. We have the following expressions,
${{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+\cdots +{}^{n}{{C}_{n}}{{x}^{n}}$
${{(1-x)}^{n}}={}^{n}{{C}_{0}}-{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}-\cdots +{{(-1)}^{n}}{}^{n}{{C}_{n}}{{x}^{n}}$
Now, we will add these two expressions. Adding the above two equations, we get the following equation,
${{(1+x)}^{n}}+{{(1-x)}^{n}}=\left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+\cdots +{}^{n}{{C}_{n}}{{x}^{n}} \right)+\left( {}^{n}{{C}_{0}}-{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+\cdots +{{(-1)}^{n}}{}^{n}{{C}_{n}}{{x}^{n}} \right)$
As the expansion of ${{(1-x)}^{n}}$ has negative sign for all the odd terms, we can see that all the odd terms in the above equation cancel out with the odd terms of the expansion of ${{(1+x)}^{n}}$. So we will have the even terms two times, as shown in the following equation,
 $\begin{align}
  & {{(1+x)}^{n}}+{{(1-x)}^{n}}=\left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}{{x}^{2}}+\cdots \right)+\left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}{{x}^{2}}+\cdots \right) \\
 & =2\cdot \left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}{{x}^{2}}+\cdots \right)
\end{align}$
Now, we will substitute $x=1$ in the above equation. So we get the following,
  $\begin{align}
  & {{(1+1)}^{n}}+{{(1-1)}^{n}}=2\cdot \left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}+\cdots \right) \\
 & {{2}^{n}}+{{0}^{n}}=2\cdot \left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}+\cdots \right) \\
\end{align}$
We can see that we are left with the sum of even binomial coefficients on the right side of the equation. Simplifying the above equation, we get
$2\cdot \left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}+\cdots \right)={{2}^{n}}$
.Now, dividing both sides by 2, we have the following equation,
$\left( {}^{n}{{C}_{0}}+{}^{n}{{C}_{2}}+\cdots \right)={{2}^{n-1}}$

So, the correct answer is “Option C”.

Note: There are certain identities and standard expressions involving binomial coefficients. These are very useful while solving such questions. When combination terms are involved, it is important to understand their definitions so that we can factor out or cancel terms. It is possible to make minor mistakes in these types of questions.