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

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

Explanation

Solution

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 nCr=n!(n−r)!×r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}

Complete step-by-step answer:
In the question, the given expression is 2nCn>4nn+1{}^{2n}{{C}_{n}}>\dfrac{{{4}^{n}}}{n+1}
Now, using combination formula on LHS i.e. nCr=n!(n−r)!×r!{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}
∴LHS=2nCn=(2n)!(2n−n)!×n!\therefore LHS={}^{2n}{{C}_{n}}=\dfrac{\left( 2n \right)!}{\left( 2n-n \right)!\times n!}
=(2n)!n!×n!=\dfrac{\left( 2n \right)!}{n!\times n!} ………………………………….(i)
Here, taking n>0n>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
=[2(1)]!1!×1!=2=\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
=[2(2)]!2!×2!=\dfrac{\left[ 2\left( 2 \right) \right]!}{2!\times 2!}
=4!2!×2!=\dfrac{4!}{2!\times 2!}
=4×3×2×12×1×2×1=\dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}
=6=6
Now, we will take 3rd value i.e. n == 3 and substituting it in equation (i), then we will get

=[2(3)]!3!×3!=\dfrac{\left[ 2\left( 3 \right) \right]!}{3!\times 3!}
=6!3!×3!=\dfrac{6!}{3!\times 3!}
=6×5×4×3×2×13×2×1×3×2×1=\dfrac{6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1\times 3\times 2\times 1}
=5×4=20=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>0n>0 for finding values.
RHS =4nn+1=\dfrac{{{4}^{n}}}{n+1} ………………………(ii)
Now taking n == 1 and substituting in equation (ii), we get
=411+1=\dfrac{{{4}^{1}}}{1+1}
=42=2=\dfrac{4}{2}=2
For n =2=2 , we get
=422+1=\dfrac{{{4}^{2}}}{2+1}
=163=\dfrac{16}{3}
=5.33=5.33
For n =3=3 , we get
=433+1=\dfrac{{{4}^{3}}}{3+1}
=644=\dfrac{64}{4}
=16=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, 2nCn>4nn+1{}^{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, (2n)\left( 2n \right) should be taken as factorial and not only 2 (n!)\left( n! \right) . This is where mistakes can happen.