Question

Prove that \dfrac{{\left( {2n!} \right)}}{n} = \left\\{ {1 \times 3 \times 5......\left( {2n - 1} \right)} \right\\}{2^n}.

Answer 1

Hint: First figure out which side you can solve, like here L.H.s is already compressed, so, R.H.S can be solved to get R.H.S. This given equation can be converted to a factorial of 2n, therefore multiply and divide R.H.S by the missing term and then solve further.

Complete step-by-step answer:
Rearranging the terms \dfrac{{\left( {2n!} \right)}}{n} = {2^n}\left\\{ {1 \times 3 \times 5......\left( {2n - 1} \right)} \right\\}
R.H.S = \left\\{ {1 \times 3 \times 5......\left( {2n - 1} \right)} \right\\}{2^n}
Multiply and divide R.H.s by 2×4×6….2n
=\dfrac{{\left\\{ {1 \times 3 \times 5......\left( {2n - 1} \right)} \right\\}{2^n}}}{{2 \times 4 \times 6......2n}} \times 2 \times 4 \times 6......2n
We get,
=\dfrac{{\left\\{ {1 \times 2 \times 3 \times 4 \times 5 \times 6......\left( {2n - 1} \right)2n} \right\\}{2^n}}}{{2 \times 4 \times 6......2n}}………..(1)
$1 \times 2 \times 3 \times 4 \times 5 \times 6......\left( {2n - 1} \right)2n$ can be written as 2n!
∴Equation (1) can be written as
=$\dfrac{{\left( {2n} \right)!{2^n}}}{{2 \times 4 \times 6......2n}}$
=$\dfrac{{\left( {2n} \right)!{2^n}}}{{{2^n}\left( {1 \times 2 \times 3....n} \right)}}$
=$\dfrac{{\left( {2n} \right)!}}{{n!}}$ = L.H.S

Hence proved.

Note: Factorials are very simple things. They are just products, indicated by an exclamation mark. In general, n! can be written as a product of all the whole numbers from 1 to n. Another use of factorial is to count how many ways you can choose things from a collection of things (in probability).