Question

Show that the middle term in the expansion of ${\left( {1 + x} \right)^{2n}}\;$ is $\dfrac{{1.3.5{\text{ }} \ldots \left( {2n{\text{ }}-{\text{ }}1} \right) \times {2^n} \times {x^n}}}{{n!}}$, where, $n \in N$.

Answer 1

Hint- Combinations are a way to calculate the total outcomes of an event where the order of the outcomes does not matter. To calculate combinations, we will use the formula $n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.
In this question, we need to show the middle term in the expansion of the function ${\left( {1 + x} \right)^{2n}}\;$is $\dfrac{{1.3.5{\text{ }} \ldots \left( {2n{\text{ }}-{\text{ }}1} \right) \times {2^n} \times {x^n}}}{{n!}}$ for which we need to carry out the formula for the binomial expansion involving the combinations as well.

Complete step by step solution:
Consider the middle term in the expansion of ${\left( {1 + x} \right)^{2n}}\;$ be ${t_{n{\text{ }} + {\text{ }}1}}$

$${t_{n{\text{ }} + {\text{ }}1}} = {}^{2n}{C_n} \times {1^{\left( {2n - n} \right)}} \times {x^n} \\\ = \dfrac{{(2n)!}}{{n!\left( {2n - n} \right)!}} \times {x^n} \\\$$

Here, we can expand the terms inside the factorials as:
${t_{n + 1}} = \dfrac{{2n\left( {{\text{ }}2n{\text{ }}-{\text{ }}1} \right)\left( {{\text{ }}2n{\text{ }}-{\text{ }}2} \right) \ldots \ldots \ldots 4 \times 3 \times 2 \times 1}}{{n!\left( n \right)!}} \times \;{x^n}$
Now, the numerator can be seen as the product of the even terms and the odd terms so, segregate the even and odd terms as:

$${t_{n + 1}} = \dfrac{{{2^n}\left[ {n\left( {n - 1} \right)\left( {n - 2} \right) \ldots ..{\text{ }} \times 2 \times 1} \right]\left[ {\left( {2n - 3} \right) \ldots \ldots 3 \times 1} \right]}}{{(n!)(n!)}} \times {\text{ }}{x^n} \\\ = \dfrac{{\left( {2n - 1} \right)\left( {2n - 3} \right) \ldots ..{\text{ }}3 \times 1}}{{n!}} \times {2^n} \times {x^n} \\\ = \dfrac{{1.3.5{\text{ }} \ldots \left( {2n{\text{ }}-{\text{ }}1} \right) \times {2^n} \times {x^n}}}{{n!}} \\\$$

Hence, the middle term in the expansion of ${\left( {1 + x} \right)^{2n}}\;$ is $\dfrac{{1.3.5{\text{ }} \ldots \left( {2n{\text{ }}-{\text{ }}1} \right) \times {2^n} \times {x^n}}}{{n!}}$.

Note: Students must be aware while taking the common terms out of the functions. Alternatively, a short method to find the value of ${}^n{C_r}$ is $n{C_r} = \dfrac{{n(n - 1)(n - 2).......(n - r + 1)}}{{r!}}$.
The above trick can be easily proved by doing simple calculations in the original equation.