Question

Let \[E = \left( {2n + 1} \right)\left( {2n + 3} \right)\left( {2n + 5} \right)...\left( {4n - 3} \right)\left( {4n - 1} \right)\] where \[n > 1\] . Then \[{2^n}E\] is divisible by
A. \[{}^n{C_{\dfrac{n}{2}}}\]
B. \[{}^{2n}{C_n}\]
C. \[{}^{3n}{C_n}\]
D. \[{}^{4n}{C_{2n}}\]

Answer 1

Hint: First, simplify the given equation by multiplying and dividing the right-hand side by \[\left( {2n} \right)!\], and \[\left( {2n + 2} \right)\left( {2n + 4} \right)\left( {2n + 6} \right)...\left( {4n - 2} \right)\left( {4n} \right)\]. Then, simplify the equation in terms of the factorial. After that, further simplify the equation by multiplying the numerator and denominator by \[n!\]. In the end, multiply both sides by \[{2^n}\] and solve it to get the required answer.

Formula Used: \[n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times .... \times 3 \times 2 \times 1\]
\[n! = n\left( {n - 1} \right)!\]

Complete step by step solution: The given equation is \[E = \left( {2n + 1} \right)\left( {2n + 3} \right)\left( {2n + 5} \right)...\left( {4n - 3} \right)\left( {4n - 1} \right)\] where \[n > 1\].

Let’s simplify the given equation.
Multiply and divide the right-hand side by \[\left( {2n} \right)!\] and \[\left( {2n + 2} \right)\left( {2n + 4} \right)\left( {2n + 6} \right)...\left( {4n - 2} \right)\left( {4n} \right)\]
We get,
\[E = \dfrac{{\left( {2n} \right)!\left( {2n + 1} \right)\left( {2n + 2} \right)\left( {2n + 3} \right)...\left( {4n - 2} \right)\left( {4n - 1} \right)\left( {4n} \right)}}{{\left( {2n} \right)!\left( {2n + 2} \right)\left( {2n + 4} \right)\left( {2n + 6} \right)...\left( {4n - 2} \right)\left( {4n} \right)}}\]
Use the factorial property \[n! = n\left( {n - 1} \right)!\] in the numerator.
\[E = \dfrac{{\left( {4n} \right)!}}{{\left( {2n} \right)!\left( {2n + 2} \right)\left( {2n + 4} \right)\left( {2n + 6} \right)...\left( {4n - 2} \right)\left( {4n} \right)}}\]
\[ \Rightarrow E = \dfrac{{\left( {4n} \right)!}}{{\left( {2n} \right)!{2^n}\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)...\left( {2n - 1} \right)\left( {2n} \right)}}\]
Again, multiply and divide the right-hand side by \[n!\].
We get,
\[E = \dfrac{{n!\left( {4n} \right)!}}{{{2^n}\left( {2n} \right)!n!\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)...\left( {2n - 1} \right)\left( {2n} \right)}}\]
Use the factorial property \[n! = n\left( {n - 1} \right)!\] in the denominator.
\[E = \dfrac{{n!\left( {4n} \right)!}}{{{2^n}\left( {2n} \right)!\left( {2n} \right)!}}\]

Now substitute the value of the above equation in \[{2^n}E\].
\[{2^n}E = {2^n}\left( {\dfrac{{n!\left( {4n} \right)!}}{{{2^n}\left( {2n} \right)!\left( {2n} \right)!}}} \right)\]
\[ \Rightarrow {2^n}E = \left( {\dfrac{{\left( {4n} \right)!}}{{\left( {2n} \right)!\left( {2n} \right)!}}} \right)n!\]
\[ \Rightarrow {2^n}E = \left( {\dfrac{{\left( {4n} \right)!}}{{\left( {2n} \right)!\left( {4n - 2n} \right)!}}} \right)n!\]
\[ \Rightarrow {2^n}E = n!{}^{4n}{C_{2n}}\]
From the above equation, we observe that \[{2^n}E\] is divisible by \[{}^{4n}{C_{2n}}\].


Option ‘D’ is correct

Note: The factorial of a number is a product of all whole numbers less than that number up to 1.
Factorial is used to solve the combination formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].