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