Question

Prove that ${\left( {^{2n}{C_0}} \right)^2} - {\left( {^{2n}{C_1}} \right)^2} + {\left( {^{2n}{C_2}} \right)^2} - \ldots + {\left( {^{2n}{C_{2n}}} \right)^2} = {\left( { - 1} \right)^n}\dfrac{{\left( {2n} \right)!}}{{n!n!}}$

Answer 1

We have to use the Binomial theorem to solve these types of questions. The Binomial theorem is ${\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{ + ^n}{C_1}{a^{n - 1}}b{ + ^n}{C_2}{a^{n - 2}}{b^2} + \ldots { + ^n}{C_n}{b^n}$. Expand ${\left( {x + 1} \right)^{2n}}$ and ${\left( {1 - x} \right)^{2n}}$ using the binomial theorem. Multiply both the equations of the expansion of ${\left( {x + 1} \right)^{2n}}$ and ${\left( {1 - x} \right)^{2n}}$. Use the formula of combinations, $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete Step by Step Solution:
From the question, we already know that we have to prove that –${\left( {^{2n}{C_0}} \right)^2} - {\left( {^{2n}{C_1}} \right)^2} + {\left( {^{2n}{C_2}} \right)^2} - \ldots + {\left( {^{2n}{C_{2n}}} \right)^2} = {\left( { - 1} \right)^n}\dfrac{{\left( {2n} \right)!}}{{n!n!}}$
We will not take left – hand side term or right – hand side term we will use the Binomial theorem to solve this question.
The Binomial theorem can be defined as the algebraic expansion of powers of the binomial. This theorem can also be called as Binomial expansion. According to this theorem, we can expand the polynomial ${\left( {a + b} \right)^n}$ into the sum. It is the method of expanding any expression which has been raised to any finite power. It is an algebraic expression which contains two different terms. If the expansion ${\left( {a + b} \right)^n}$ is given then, by using the Binomial theorem it can be expanded as –
${\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{ + ^n}{C_1}{a^{n - 1}}b{ + ^n}{C_2}{a^{n - 2}}{b^2} + \ldots { + ^n}{C_n}{b^n}$
In the sigma notation, the above expansion can also be written as –
$\sum\limits_{r = 0}^n {^n{C_r}{a^{n - r}}{b^r}}$
To prove the identity given in the question, we will use the Binomial Theorem to expand ${\left( {x + 1} \right)^{2n}}$ and ${\left( {1 - x} \right)^{2n}}$ . Therefore, they can be expanded as –
$\Rightarrow {\left( {x + 1} \right)^{2n}}{ = ^{2n}}{C_0}{x^{2n}}{ + ^{2n}}{C_1}{x^{2n - 1}}{ + ^{2n}}{C_2}{x^{2n - 2}} + \ldots { + ^{2n}}{C_{2n}} \cdots \left( 1 \right) \\\ \Rightarrow {\left( {1 - x} \right)^{2n}}{ = ^{2n}}{C_0}{ + ^{2n}}{C_1}x{ + ^{2n}}{C_2}{x^2} + \ldots { + ^{2n}}{C_{2n}}{x^{2n}} \cdots \left( 2 \right) \\\$
Now, multiply the equation (1) and equation (2) with each other, we get –
$\Rightarrow {\left( {x + 1} \right)^{2n}}{\left( {1 - x} \right)^{2n}} = \left( {^{2n}{C_0}{x^{2n}}{ + ^{2n}}{C_1}{x^{2n - 1}}{ + ^{2n}}{C_2}{x^{2n - 2}} + \ldots { + ^{2n}}{C_{2n}}} \right).\left( {^{2n}{C_0}{ + ^{2n}}{C_1}x{ + ^{2n}}{C_2}{x^2} + \ldots { + ^{2n}}{C_{2n}}{x^{2n}}} \right)$
The above expression in the left – hand side has same powers raised in the term, so, we get –
$\Rightarrow {\left[ {\left( {1 + x} \right)\left( {1 - x} \right)} \right]^{2n}} = \left( {^{2n}{C_0}{x^{2n}}{ + ^{2n}}{C_1}{x^{2n - 1}}{ + ^{2n}}{C_2}{x^{2n - 2}} + \ldots { + ^{2n}}{C_{2n}}} \right).\left( {^{2n}{C_0}{ + ^{2n}}{C_1}x{ + ^{2n}}{C_2}{x^2} + \ldots { + ^{2n}}{C_{2n}}{x^{2n}}} \right)$
We know the identity of differences of squares which is –
$\left( {a - b} \right)\left( {a + b} \right) = \left( {{a^2} - {b^2}} \right)$
So, by using the above identity, we can write the above expansion as –
$\Rightarrow {\left( {1 - {x^2}} \right)^{2n}} = \left( {^{2n}{C_0}{x^{2n}}{ + ^{2n}}{C_1}{x^{2n - 1}}{ + ^{2n}}{C_2}{x^{2n - 2}} + \ldots { + ^{2n}}{C_{2n}}} \right).\left( {^{2n}{C_0}{ + ^{2n}}{C_1}x{ + ^{2n}}{C_2}{x^2} + \ldots { + ^{2n}}{C_{2n}}{x^{2n}}} \right)$
In the sigma notation, the above expression can be rewritten as –
$\Rightarrow {\left( {1 - {x^2}} \right)^{2n}} = \sum\limits_{r = 0}^{2n} {^{2n}{C_r}{{\left( { - {x^2}} \right)}^r}}$
Therefore, now equating the coefficients of ${x^{2n}}$ terms, we get –
$\therefore {\left( {^{2n}{C_0}} \right)^2} - {\left( {^{2n}{C_1}} \right)^2} + {\left( {^{2n}{C_2}} \right)^2} + \ldots + {\left( {^{2n}{C_{2n}}} \right)^2} = {\left( { - 1} \right)^n}{.^{2n}}{C_n} \cdots \left( 1 \right)$
Now, we know the formula of solving the combinations which is –
$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Using the above formula in the equation (1) in the right – hand side term, we get –
$\therefore {\left( {^{2n}{C_0}} \right)^2} - {\left( {^{2n}{C_1}} \right)^2} + {\left( {^{2n}{C_2}} \right)^2} + \ldots + {\left( {^{2n}{C_{2n}}} \right)^2} = {\left( { - 1} \right)^n}.\dfrac{{2n!}}{{n!n!}} \cdots \left( 1 \right)$
Hence, this is the required proof which was required in the question.

Note:
Many students can make mistakes and take the left – hand side term or right – hand side term and try to solve this question but it cannot be solved by that method. The formula for the Binomial theorem should be remembered by students as many terms require the Binomial theorem to be expanded. So, this is an important theorem for solving questions.