Question

The value of $^{n}{{C}_{0}}^{n}{{C}_{2}}{{+}^{n}}{{C}_{1}}^{n}{{C}_{3}}{{+}^{n}}{{C}_{2}}^{n}{{C}_{4}}+.........{{+}^{n}}{{C}_{n-2}}^{n}{{C}_{n}}$ is equal to:
(a) $^{2n}{{C}_{n-2}}$
(b) $^{2n}{{C}_{n+1}}$
(c) $^{2n}{{C}_{n-1}}$
(d) None of these.

Answer 1

Hint: Notice the pattern of the expression given in the question and relate the expression with the coefficient of ${{x}^{n-2}}$ in the expansion of ${{\left( 1+x \right)}^{2n}}$ . Also, try to convert the expression to a form where you can see that the sum of lower indexes of each term is constant.

Complete step-by-step solution:
Let us start with the solution to the above question by simplifying the expression given in the question.
$^{n}{{C}_{0}}^{n}{{C}_{2}}{{+}^{n}}{{C}_{1}}^{n}{{C}_{3}}{{+}^{n}}{{C}_{2}}^{n}{{C}_{4}}+.........{{+}^{n}}{{C}_{n-2}}^{n}{{C}_{n}}$
We know that $^{n}{{C}_{r}}{{=}^{n}}{{C}_{n-r}}$ . So, we get
$^{n}{{C}_{0}}^{n}{{C}_{2}}{{+}^{n}}{{C}_{1}}^{n}{{C}_{3}}{{+}^{n}}{{C}_{2}}^{n}{{C}_{4}}+.........{{+}^{n}}{{C}_{n-2}}^{n}{{C}_{n}}{{=}^{n}}{{C}_{0}}^{n}{{C}_{n-2}}{{+}^{n}}{{C}_{1}}^{n}{{C}_{n-3}}{{+}^{n}}{{C}_{2}}^{n}{{C}_{n-4}}+.........{{+}^{n}}{{C}_{n-2}}^{n}{{C}_{0}}$The point to be noted from result is that the sum of the lower index is constant for each term.
Now, the binomial expansion of ${{\left( 1+x \right)}^{n}}$ is:
\[{{\left( 1+x \right)}^{n}}={{\text{ }}^{n}}{{\text{C}}_{0}}{{x}^{0}}{{+}^{n}}{{\text{C}}_{1}}{{x}^{1}}{{+}^{n}}{{\text{C}}_{2}}{{x}^{2}}+........{{+}^{n}}{{\text{C}}_{n}}{{x}^{n}}=\sum{^{n}{{C}_{r}}{{x}^{r}}}\]
Using the above expansion in the summation form, we can say that:
\[{{\left( 1+x \right)}^{n}}{{\left( 1+x \right)}^{n}}=\sum\limits_{k}{\sum\limits_{r}{^{n}{{C}_{r}}^{n}{{C}_{k}}{{x}^{r}}{{x}^{k}}}}\]
\[\Rightarrow {{\left( 1+x \right)}^{2n}}=\sum\limits_{k}{\sum\limits_{r}{^{n}{{C}_{r}}^{n}{{C}_{k}}{{x}^{r+k}}}}\]
So, we can find the coefficient of ${{x}^{n-2}}$ in the by using the constraint that r+k=n and both r and k are whole numbers less than n-2. As from the simplified expression it is clear that the terms are the coefficient of ${{x}^{n-2}}$ , as the sum of the lower indices is n-2.
\[\text{coefficient of }{{\text{x}}^{n-2}}\text{ in the expansion of }{{\left( 1+x \right)}^{2n}}=\]$^{n}{{C}_{0}}^{n}{{C}_{2}}{{+}^{n}}{{C}_{1}}^{n}{{C}_{3}}+.........{{+}^{n}}{{C}_{n-2}}^{n}{{C}_{n}}$
Now according to the binomial theorem, we know that\[\text{coefficient of }{{\text{x}}^{n-2}}\text{ in the expansion of }{{\left( 1+x \right)}^{2n}}{{=}^{2n}}{{C}_{n-2}}\]
\[{{\therefore }^{2n}}{{C}_{n-2}}{{=}^{n}}{{C}_{0}}^{n}{{C}_{2}}{{+}^{n}}{{C}_{1}}^{n}{{C}_{3}}+.........{{+}^{n}}{{C}_{n-2}}^{n}{{C}_{n}}\]
Now the right side of the equation is the expression given in the question. Therefore, the answer to the above question is option (A).

Note: Always be careful with the signs that appear in the expansions, as the students are generally finding signs to be a concern while using the binomial expansions. Also, be careful about the calculation part, as in general cases, the questions involving binomial expansion contain very long and complex calculations due to the presence of factorial terms. You should also know that the binomial coefficient and actual coefficients might or might not be the same. For example: in the expansion of ${{\left( 1+3x \right)}^{3}}$ , the binomial coefficient of ${{x}^{3}}$ is $^{3}{{C}_{3}}=1$ and coefficient is 27.