Question

Show that the coefficient of middle term of ${{\left( 1+x \right)}^{2n}}$ is equal to the sum of coefficient of the two middle term of ${{\left( 1+x \right)}^{2n-1}}$

Answer 1

Hint: Try to use the binomial expansion formula. Note that the middle term of ‘m’ numbers is
$\dfrac{m+1}{2}$ if m is odd, or $\dfrac{m}{2}$ and $\dfrac{m}{2}+1$ if m is even.

Complete step-by-step answer:
We know that by binomial formula,
${{\left( 1+x \right)}^{m}}=C_{0}^{m}{{x}^{m}}+C_{1}^{m}{{x}^{m-1}}+......+C_{m}^{m}{{x}^{0}}.....\left( 1 \right)$

Therefore applying this to the given expression ${{\left( 1+x \right)}^{2n}}$, we get
${{\left( 1+x \right)}^{2n}}=C_{0}^{2n}{{x}^{2n}}+C_{1}^{2n}{{x}^{2n-1}}+......+C_{2n}^{2n}{{x}^{0}}.....\left( 2 \right)$

Also, number of terms in ‘m’ power is m+1

Therefore, the total number of terms = 2n+1 i.e., odd number.
So, the middle term is given by, ${{\left( \dfrac{2n+1+1}{2} \right)}^{th}}$ term.
That is,
${{\left( \dfrac{2(n+1)}{2} \right)}^{th}}={{(n+1)}^{th}}$
So, the middle term of ${{\left( 1+x \right)}^{2n}}$ is the (n+1)th term. It is given by,
$C_{n}^{2n}{{x}^{n}}............\left( 3 \right)$

We know the coefficient of the kth term is given by $C_{k-1}^{2n}$.
Therefore the coefficient of the (n+1)th is,
$C_{n}^{2n}=\dfrac{\left( 2n \right)!}{\left( 2n-n \right)!\left( n \right)!}..........\left( 4 \right)$

Now consider the expansion of ${{\left( 1+x \right)}^{2n-1}}$ .
It contains (2n-1+1)=2n terms, this is an even number.

Therefore, middle terms of even numbers are n and (n+1)th terms.
So, the middle terms of ${{\left( 1+x \right)}^{2n-1}}$ are given by,
nth term: $C_{n-1}^{2n-1}{{x}^{n+1}}$
(n+1)th term: $C_{n}^{2n-1}{{x}^{n}}$
Now the coefficient of nth term =$C_{n-1}^{2n-1}.........\left( 5 \right)$
Coefficient of (n+1)th term = $C_{n}^{2n-1}.........\left( 6 \right)$

Adding equation (5) and (6), we get
\[C_{n-1}^{2n-1}+C_{n}^{2n-1}=\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!n!}+\dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!n!}........\left( 7 \right)\]

Here we have used the formula,
$C_{i}^{m}=\dfrac{m!}{\left( m-1 \right)i!}$ where $i!=i\left( i-1 \right)\left( i-2 \right).....1$
Solving equation (7), we get
$C_{n-1}^{2n-1}+C_{n}^{2n-1}=2\left( \dfrac{\left( 2n-1 \right)!}{\left( n-1 \right)!n!} \right)$

Multiply numerator and denominator by n, the above equation can be written as
$C_{n-1}^{2n-1}+C_{n}^{2n-1}=\dfrac{2n\left( 2n-1 \right)!}{n\left( n-1 \right)!n!}$
From the definition of ‘!’ (Factorial) we have, 2n(2n-1)! = (2n)! and n(n-1)! = n!, so the above equation can be written as
$C_{n-1}^{2n-1}+C_{n}^{2n-1}=\dfrac{\left( 2n \right)!}{n!n!}$
Now ‘n’ can be written as ‘2n-n’, so the above equation can be written as,
$C_{n-1}^{2n-1}+C_{n}^{2n-1}=\dfrac{\left( 2n \right)!}{\left( 2n-n \right)!\left( n \right)!}$

Comparing this with equation (4), we get
$C_{n-1}^{2n-1}+C_{n}^{2n-1}=C_{n}^{2n}$
So, the coefficient of middle term of ${{\left( 1+x \right)}^{2n}}$ is equal to the sum of coefficients of the two middle terms of ${{\left( 1+x \right)}^{2n-1}}$.
Hence, proved

Note: Students often think of power as a number of terms. This is false, for example, \[{{\left( 1+x \right)}^{0}}\] it has 1 term but the power is 0. ${{\left( 1+x \right)}^{1}}$has 2 terms and not 1 term.
Also, the kth term is not $C_{k}^{m}{{I}^{k}}$ , but $C_{2-1}^{m}{{I}^{k}}$.