# 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 ${{\left( 1+x \right)}^{2n-1}}$

Last updated date: 23rd Mar 2023

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Answer

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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 $C_{2-1}^{m}{{I}^{k}}$.

$\dfrac{m+1}{2}$ 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 $C_{2-1}^{m}{{I}^{k}}$.

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