Question

Prove that \[{}^{2n}{{C}_{n}}>\dfrac{{{4}^{n}}}{n+1}\] using proper method.

Answer 1

Hint: Here, we have to prove the given expression by taking left hand side (LHS) and right hand side (RHS) separately and finding values of it. Also, will use formula of combination \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}\]

Complete step-by-step answer:
In the question, the given expression is \[{}^{2n}{{C}_{n}}>\dfrac{{{4}^{n}}}{n+1}\]
Now, using combination formula on LHS i.e. \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}\]
\[\therefore LHS={}^{2n}{{C}_{n}}=\dfrac{\left( 2n \right)!}{\left( 2n-n \right)!\times n!}\]
\[=\dfrac{\left( 2n \right)!}{n!\times n!}\] ………………………………….(i)
Here, taking \[n>0\] and substituting in equation (i) to find the values. So, we will first take 1st value i.e. n \[=\] 1 and substituting it in equation (i), then we will get
\[=\dfrac{\left[ 2\left( 1 \right) \right]!}{1!\times 1!}=2\]
Now, we will take 2nd value i.e. n \[=\] 2 and substituting it in equation (i), then we will get
\[=\dfrac{\left[ 2\left( 2 \right) \right]!}{2!\times 2!}\]
\[=\dfrac{4!}{2!\times 2!}\]
\[=\dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}\]
\[=6\]
Now, we will take 3rd value i.e. n \[=\] 3 and substituting it in equation (i), then we will get

\[=\dfrac{\left[ 2\left( 3 \right) \right]!}{3!\times 3!}\]
\[=\dfrac{6!}{3!\times 3!}\]
\[=\dfrac{6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1\times 3\times 2\times 1}\]
\[=5\times 4=20\]
So, for n \[=\] 1,2,3, …. We get values for LHS as 2,6,20, ….so on.
Now, taking right hand side (RHS) and substituting values of \[n>0\] for finding values.
RHS \[=\dfrac{{{4}^{n}}}{n+1}\] ………………………(ii)
Now taking n \[=\] 1 and substituting in equation (ii), we get
\[=\dfrac{{{4}^{1}}}{1+1}\]
\[=\dfrac{4}{2}=2\]
For n \[=2\] , we get
\[=\dfrac{{{4}^{2}}}{2+1}\]
\[=\dfrac{16}{3}\]
\[=5.33\]
For n \[=3\] , we get
\[=\dfrac{{{4}^{3}}}{3+1}\]
\[=\dfrac{64}{4}\]
\[=16\]
So, for n \[=\] 1,2,3, …. We get values for RHS as 2, 5.33, 16, …… so on.
Comparing values obtained of LHS and RHS and we can see that
\[\left\{ 2,6,20,...... \right\}>\left\{ 2,5.33,16,...... \right\}\]
So, LHS \[>\] RHS
 Therefore, \[{}^{2n}{{C}_{n}}>\dfrac{{{4}^{n}}}{n+1}\]
Hence, proved.

Note: Students should be careful while substituting values in the LHS side as first it needs to be solved using appropriate formulas. Also, $\left( 2n \right)$ should be taken as factorial and not only 2 $\left( n! \right)$ . This is where mistakes can happen.