Question

Prove that: \[{}^{2n}{c_n} = \dfrac{{{2^n}\left\{ {1.3.5............\left( {2n - 1} \right)} \right\}}}{{n!}}\]

Answer 1

Hint: To solve this question first we have to use the formula for expanding the term in factorial. Then expand that part to the multiplication of numbers. Then the common terms that are possible to take are common and try to form the factorial part and cancel that part from the numerator and denominator then we are able to find that same term which we have to prove.

Complete step-by-step solution:
First we have to expand the given part in factorial.
use the formula of \[{}^{2n}{c_n} = \dfrac{{\left( {2n} \right)!}}{{n!n!}}\]
on expanding this terms in a numbers
\[\dfrac{{\left( {2n} \right)!}}{{n!n!}} = \dfrac{{2n\left( {2n - 1} \right)\left( {2n - 2} \right)\left( {2n - 3} \right)\left( {2n - 4} \right)................3.2.1}}{{n!n!}}\]
On taking \[{2^n}\] common from the terms which are divisible by 2 because we are able to find the \[{2^n}\]term in the final answer.
\[\dfrac{{\left( {2n} \right)!}}{{n!n!}} = \dfrac{{{2^n}\left( {n\left( {2n - 1} \right)\left( {n - 1} \right)\left( {2n - 3} \right)\left( {n - 2} \right)................3.1.1} \right)}}{{n!n!}}\]
In this expression we found an expression of \[n!\].
So taking that expression out from the given expression.
\[\dfrac{{\left( {2n} \right)!}}{{n!n!}} = \dfrac{{{2^n}n!\left( {\left( {2n - 1} \right)\left( {2n - 3} \right)................3.1} \right)}}{{n!n!}}\]
On cancelling \[n!\] apart from numerator and denominator.
\[\dfrac{{\left( {2n} \right)!}}{{n!n!}} = \dfrac{{{2^n}\left( {\left( {2n - 1} \right)\left( {2n - 3} \right)................3.1} \right)}}{{n!}}\]
On looking at this expression we found that we have to prove.
\[\dfrac{{\left( {2n} \right)!}}{{n!n!}} = \dfrac{{{2^n}\left( {\left( {2n - 1} \right)\left( {2n - 3} \right)................3.1} \right)}}{{n!}}\]
Hence proved.

Note: Students easily expand the given term but they are unable to take the common part out from the whole expression they stuck there and are unable to form the \[n!\] part in the numerator and they are unable to cancel out \[n!\] part and unable to prove this part. To solve this type of question you must have a good knowledge of the factorial part and you must have a good practice to take some common. Like this, there are many more conditions which we can prove by the same method that is taking common and canceling the common part from numerator and denominator.