Solveeit Logo
Login

Question

Question: Show that the coefficient of middle term of \({{\left( 1+x \right)}^{2n}}\) is equal to the sum of c...

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

Explanation

Solution

Hint: Try to use the binomial expansion formula. Note that the middle term of ‘m’ numbers is
m+12\dfrac{m+1}{2} if m is odd, or m2\dfrac{m}{2} and m2+1\dfrac{m}{2}+1 if m is even.

Complete step-by-step answer:
We know that by binomial formula,
(1+x)m=C0mxm+C1mxm−1+......+Cmmx0.....(1){{\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 (1+x)2n{{\left( 1+x \right)}^{2n}}, we get
(1+x)2n=C02nx2n+C12nx2n−1+......+C2n2nx0.....(2){{\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, (2n+1+12)th{{\left( \dfrac{2n+1+1}{2} \right)}^{th}} term.
That is,
(2(n+1)2)th=(n+1)th{{\left( \dfrac{2(n+1)}{2} \right)}^{th}}={{(n+1)}^{th}}
So, the middle term of (1+x)2n{{\left( 1+x \right)}^{2n}} is the (n+1)th term. It is given by,
Cn2nxn............(3)C_{n}^{2n}{{x}^{n}}............\left( 3 \right)

We know the coefficient of the kth term is given by Ck−12nC_{k-1}^{2n}.
Therefore the coefficient of the (n+1)th is,
Cn2n=(2n)!(2n−n)!(n)!..........(4)C_{n}^{2n}=\dfrac{\left( 2n \right)!}{\left( 2n-n \right)!\left( n \right)!}..........\left( 4 \right)

Now consider the expansion of (1+x)2n−1{{\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 (1+x)2n−1{{\left( 1+x \right)}^{2n-1}} are given by,
nth term: Cn−12n−1xn+1C_{n-1}^{2n-1}{{x}^{n+1}}
(n+1)th term: Cn2n−1xnC_{n}^{2n-1}{{x}^{n}}
Now the coefficient of nth term =Cn−12n−1.........(5)C_{n-1}^{2n-1}.........\left( 5 \right)
Coefficient of (n+1)th term = Cn2n−1.........(6)C_{n}^{2n-1}.........\left( 6 \right)

Adding equation (5) and (6), we get
Cn−12n−1+Cn2n−1=(2n−1)!(n−1)!n!+(2n−1)!(n−1)!n!........(7)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,
Cim=m!(m−1)i!C_{i}^{m}=\dfrac{m!}{\left( m-1 \right)i!} where i!=i(i−1)(i−2).....1i!=i\left( i-1 \right)\left( i-2 \right).....1
Solving equation (7), we get
Cn−12n−1+Cn2n−1=2((2n−1)!(n−1)!n!)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
Cn−12n−1+Cn2n−1=2n(2n−1)!n(n−1)!n!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
Cn−12n−1+Cn2n−1=(2n)!n!n!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,
Cn−12n−1+Cn2n−1=(2n)!(2n−n)!(n)!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
Cn−12n−1+Cn2n−1=Cn2nC_{n-1}^{2n-1}+C_{n}^{2n-1}=C_{n}^{2n}
So, the coefficient of middle term of (1+x)2n{{\left( 1+x \right)}^{2n}} is equal to the sum of coefficients of the two middle terms of (1+x)2n−1{{\left( 1+x \right)}^{2n-1}}.
Hence, proved

Note: Students often think of power as a number of terms. This is false, for example, (1+x)0{{\left( 1+x \right)}^{0}} it has 1 term but the power is 0. (1+x)1{{\left( 1+x \right)}^{1}}has 2 terms and not 1 term.
Also, the kth term is not CkmIkC_{k}^{m}{{I}^{k}} , but C2−1mIkC_{2-1}^{m}{{I}^{k}}.