Question

If we have an expression as ${{\left( 1+x \right)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+...+{{C}_{n}}{{x}^{n}}$, then prove the following: ${{C}_{0}}{{C}_{r}}+{{C}_{1}}{{C}_{r+1}}+...+{{C}_{n-r}}{{C}_{n}}=\dfrac{\left( 2n \right)!}{\left( n+r \right)!\left( n-r \right)!}$

Answer 1

To solve this question, firstly we will start with following series$^{n}{{C}_{0}}{}^{n}{{C}_{r}}+{}^{n}{{C}_{1}}{}^{n}{{C}_{r+1}}+...+{}^{n}{{C}_{n-r}}{}^{n}{{C}_{n}}$ which is equals to the coefficient of ${{x}^{n+r}}$ in the expansion of ${{\left( 1+x \right)}^{n}}\cdot {{\left( x+1 \right)}^{n}}$ . Now we will equate the coefficient with the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1+x \right)}^{2n}}$ . Now substituting r = n we get the required equation.

Complete step-by-step solution:
In this question, we are given that ${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+...+{}^{n}{{C}_{n}}{{x}^{n}}$
Using this information, we need to prove that $^{n}{{C}_{0}}{}^{n}{{C}_{r}}+{}^{n}{{C}_{1}}{}^{n}{{C}_{r+1}}+...+{}^{n}{{C}_{n-r}}{}^{n}{{C}_{n}}=\dfrac{\left( 2n \right)!}{\left( n+r \right)!\left( n-r \right)!}$.
The given sequence is a binomial sequence. Let us first define the binomial theorem.
In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial ${{\left( x+y \right)}^{n}}$ into a sum involving terms of the form $a{{x}^{b}}{{y}^{c}}$, where the exponents b and c are non negative integers with b + c = n, and the coefficient a of each term is a specific positive integer depending on n and b.
Now consider ${{\left( 1+x \right)}^{n}}\cdot {{\left( x+1 \right)}^{n}}$
Let us find the coefficient of ${{x}^{n+r}}$ in the equation.
Now ${{x}^{r}}$ can be formed by taking x in ${{\left( 1+x \right)}^{n}}$ and ${{x}^{n}}$ from ${{\left( x+1 \right)}^{n}}$
or by taking ${{x}^{r-1}}$ in ${{\left( 1+x \right)}^{n}}$ and ${{x}^{n+1}}$ from the term ${{\left( x+1 \right)}^{n}}$ .
Hence taking all the coefficients we get the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1+x \right)}^{n}}{{\left( 1+x \right)}^{n}}$ is given by $^{n}{{C}_{0}}{}^{n}{{C}_{r}}+{}^{n}{{C}_{1}}{}^{n}{{C}_{r+1}}+...+{}^{n}{{C}_{n-r}}{}^{n}{{C}_{n}}$
Now we know that ${{\left( 1+x \right)}^{n}}{{\left( 1+x \right)}^{n}}={{\left( 1+x \right)}^{2n}}$
Now let us find the coefficient of ${{x}^{n+r}}$ in the expansion of ${{\left( 1+x \right)}^{2n}}$ by using binomial theorem. Hence we get the coefficient as ${}^{2n}{{C}_{n+r}}$
Hence we can say that ${}^{2n}{{C}_{r+1}}{{=}^{n}}{{C}_{0}}{}^{n}{{C}_{r}}+{}^{n}{{C}_{1}}{}^{n}{{C}_{n+r}}+...+{}^{n}{{C}_{n-r}}{}^{n}{{C}_{n}}$
Now simplifying the equation we get,
$\begin{aligned} & \Rightarrow \dfrac{\left( 2n \right)!}{\left( 2n-n-r \right)!\left( n+r \right)!}{{=}^{n}}{{C}_{0}}{}^{n}{{C}_{r}}+{}^{n}{{C}_{1}}{}^{n}{{C}_{r+1}}+...+{}^{n}{{C}_{n-r}}{}^{n}{{C}_{n}} \\\ & \Rightarrow \dfrac{\left( 2n \right)!}{\left( n-r \right)!\left( n+r \right)!}{{=}^{n}}{{C}_{0}}{}^{n}{{C}_{r}}+{}^{n}{{C}_{1}}{}^{n}{{C}_{r+1}}+...+{}^{n}{{C}_{n-r}}{}^{n}{{C}_{n}} \\\ \end{aligned}$
Hence the required equation is proved.

Note: In this question, it is very important to note that in the expansion given in the question or those used in the solution, the terms used for coefficients are ${{C}_{i}}$. This ${{C}_{i}}$ is equal to $^{n}{{C}_{i}}=\dfrac{n!}{i!\left( n-i \right)!}$, which is equals to the coefficient of ${{x}^{i}}$ in the binomial expansion of ${{\left( 1+x \right)}^{n}}$.