Question

The value of ${}^{n}{{C}_{0}}{}^{n}{{C}_{r}}+{}^{n}{{C}_{1}}{}^{n}{{C}_{r+1}}+{}^{n}{{C}_{2}}{}^{n}{{C}_{r+2}}+....+{}^{n}{{C}_{n-r}}{}^{n}{{C}_{n}}$ is equal to?
(a) ${}^{2n}{{C}_{r}}$
(b) ${}^{2n}{{C}_{n+r}}$
(c) ${}^{2n}{{C}_{r-1}}$
(d) ${}^{2n}{{C}_{r+1}}$

Answer 1

Use the binomial expansion of the expression ${{\left( 1+x \right)}^{n}}$ given as ${{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{0}}+{}^{n}{{C}_{1}}{{x}^{1}}+{}^{n}{{C}_{2}}{{x}^{2}}+....+{}^{n}{{C}_{n}}{{x}^{n}}$ and assume it as equation (i). Now, interchange the place of 1 and x in the expression ${{\left( 1+x \right)}^{n}}$ and then use the same binomial expansion to write ${{\left( x+1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}+....+{}^{n}{{C}_{n}}{{x}^{0}}$ and assume it as expression (ii). Multiply equation (i) and (ii) and check the terms of x in which we will get the coefficients ${}^{n}{{C}_{0}}{}^{n}{{C}_{r}}+{}^{n}{{C}_{1}}{}^{n}{{C}_{r+1}}+{}^{n}{{C}_{2}}{}^{n}{{C}_{r+2}}+....+{}^{n}{{C}_{n-r}}{}^{n}{{C}_{n}}$. Write the formula for the coefficient of the general term of ${{\left( 1+x \right)}^{2n}}$ given as ${{T}_{a+1}}={}^{2n}{{C}_{a}}$ and substitute the suitable value of a in terms of n and r. Use the property ${}^{p}{{C}_{q}}={}^{p}{{C}_{p-q}}$ to get the answer.

Complete step-by-step solution:
Here we have been provided with the expression ${}^{n}{{C}_{0}}{}^{n}{{C}_{r}}+{}^{n}{{C}_{1}}{}^{n}{{C}_{r+1}}+{}^{n}{{C}_{2}}{}^{n}{{C}_{r+2}}+....+{}^{n}{{C}_{n-r}}{}^{n}{{C}_{n}}$ and we are asked to find its value. Let us use the binomial expansion of ${{\left( 1+x \right)}^{n}}$ in different manners to get the answer.
Now, we know that the expansion formula of the binomial expression ${{\left( 1+x \right)}^{n}}$ is given as: -
$\Rightarrow {{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{0}}+{}^{n}{{C}_{1}}{{x}^{1}}+{}^{n}{{C}_{2}}{{x}^{2}}+....+{}^{n}{{C}_{n}}{{x}^{n}}$ ………. (i)
Interchanging the terms 1 and x in the L.H.S of the above expression and then using the same formula we get,
$\Rightarrow {{\left( x+1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}+....+{}^{n}{{C}_{n}}{{x}^{0}}$
$\Rightarrow {{\left( x+1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}+{}^{n}{{C}_{r}}{{x}^{n-r}}+{}^{n}{{C}_{r+1}}{{x}^{n-\left( r+1 \right)}}....+{}^{n}{{C}_{n}}{{x}^{0}}$ ……… (ii)
Multiplying equations (i) and (ii) we will see that we will get the coefficients ${}^{n}{{C}_{0}}{}^{n}{{C}_{r}},{}^{n}{{C}_{1}}{}^{n}{{C}_{r+1}},{}^{n}{{C}_{2}}{}^{n}{{C}_{r+2}},....,{}^{n}{{C}_{n-r}}{}^{n}{{C}_{n}}$ in terms containing ${{x}^{n-r}}$. For example: - the term ${}^{n}{{C}_{0}}$ of equation (i) will be multiplied with the term containing ${}^{n}{{C}_{r}}$ of equation (ii) and so on the process will go forward for the other terms required. Therefore, the required sum will be equal to the coefficient of the term containing ${{x}^{n-r}}$ in the product ${{\left( 1+x \right)}^{n}}\times {{\left( x+1 \right)}^{n}}={{\left( 1+x \right)}^{2n}}$.
The general term of the expression ${{\left( 1+x \right)}^{2n}}$ is given as${{T}_{a+1}}={}^{2n}{{C}_{a}}{{x}^{a}}$, so substituting a = n – r in the formula we get,
$\Rightarrow$ Coefficient of ${{x}^{n-r}}={}^{2n}{{C}_{n-r}}$
Using the formula ${}^{p}{{C}_{q}}={}^{p}{{C}_{p-q}}$ we get,
$\Rightarrow$ Coefficient of ${{x}^{n-r}}={}^{2n}{{C}_{2n-\left( n-r \right)}}$
$\therefore$ Coefficient of ${{x}^{n-r}}={}^{2n}{{C}_{n+r}}$
Hence, option (b) is the correct answer.

Note: Remember some basic expansion formulas of the binomial expressions like ${{\left( 1+x \right)}^{n}}$ and ${{\left( x+y \right)}^{n}}$ as they are often used in the topic ‘Binomial Theorem’. Note that the formula ${}^{p}{{C}_{q}}={}^{p}{{C}_{p-q}}$ can be proved by using the general formula of the expression ${}^{p}{{C}_{q}}$ given as $\dfrac{p!}{q!\left( p-q \right)!}$. Just substitute p – q in place of q to get the required result.