# Show that the middle term in the expansion of ${\left( {1 + x} \right)^n}$is $6{x^2}$ if n = 4.

Last updated date: 14th Mar 2023

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Answer

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Hint - There can be two methods to solve this problem, one is based upon the general formula of expansion of ${\left( {1 + x} \right)^n}$ using the binomial expansion and the other one focuses on direct formula to find the middle term in expansion of ${\left( {1 + x} \right)^n}$ depending upon whether n is even or odd.

Now let’s use the direct formula for finding the middle term in the expansion of ${\left( {1 + x} \right)^n}$.

Here n =4 (given in question)

Clearly n is even thus the middle term in expansion of ${\left( {1 + x} \right)^n}$is $^n{C_{\dfrac{n}{2}}}{x^{\dfrac{n}{2}}}$……………… (1)

So let’s directly put the value of n in equation (1) we get

Middle term will be $^4{C_{\dfrac{4}{2}}}{x^{\dfrac{4}{2}}} = {{\text{ }}^4}{C_2}{x^2}$………………….. (2)

Now The formula for $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$…………………. (3)

Using equation three we get

$

^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} \\

\Rightarrow \dfrac{{4!}}{{2!\left( 2 \right)!}} \\

$

Now $n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right).........\left( {n - r} \right)$ such that $r < n$

Using this concept we get

$

^4{C_2} = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} \\

\Rightarrow 6 \\

$

Hence putting it in equation (2) we get

Middle term of ${\left( {1 + x} \right)^n}$is $6{x^2}$

Note – Now let’s talk about a method in which we can use the entire expansion of ${\left( {1 + x} \right)^n}$using binomial theorem. The expansion of ${\left( {1 + x} \right)^n}$is$1 + nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ............\infty $. Hence after completely expanding till the given value of n, find the middle term in that series, this will also give the right answer.

Now let’s use the direct formula for finding the middle term in the expansion of ${\left( {1 + x} \right)^n}$.

Here n =4 (given in question)

Clearly n is even thus the middle term in expansion of ${\left( {1 + x} \right)^n}$is $^n{C_{\dfrac{n}{2}}}{x^{\dfrac{n}{2}}}$……………… (1)

So let’s directly put the value of n in equation (1) we get

Middle term will be $^4{C_{\dfrac{4}{2}}}{x^{\dfrac{4}{2}}} = {{\text{ }}^4}{C_2}{x^2}$………………….. (2)

Now The formula for $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$…………………. (3)

Using equation three we get

$

^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}} \\

\Rightarrow \dfrac{{4!}}{{2!\left( 2 \right)!}} \\

$

Now $n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right).........\left( {n - r} \right)$ such that $r < n$

Using this concept we get

$

^4{C_2} = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} \\

\Rightarrow 6 \\

$

Hence putting it in equation (2) we get

Middle term of ${\left( {1 + x} \right)^n}$is $6{x^2}$

Note – Now let’s talk about a method in which we can use the entire expansion of ${\left( {1 + x} \right)^n}$using binomial theorem. The expansion of ${\left( {1 + x} \right)^n}$is$1 + nx + \dfrac{{n\left( {n - 1} \right)}}{2}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ............\infty $. Hence after completely expanding till the given value of n, find the middle term in that series, this will also give the right answer.

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