Question

If n=10 then${{C}_{0}}^{2}-{{C}_{1}}^{2}+{{C}_{2}}^{2}-{{C}_{3}}^{2}...........{{\left( -1 \right)}^{n}}{{C}_{n}}^{2}$ equals
a) ${{\left( -1 \right)}^{5}}{}^{10}{{C}_{5}}$
b) 0
c) ${}^{10}{{C}_{5}}$
d) ${{\left( -1 \right)}^{6}}{}^{10}{{C}_{6}}$

Answer 1

We know the binomial expansion of ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}.........{}^{n}{{C}_{n}}{{b}^{n}}$ We will write the binomial expansion for ${{\left( 1+x \right)}^{n}}$and binomial expansion for ${{\left( x-1 \right)}^{n}}$ we will now multiply both the expansion we will get a expression with coefficient as ${{x}^{n}}$. We will get ${{\left( 1+x \right)}^{n}}{{\left( x-1 \right)}^{n}}$ on the left-hand side of the equation.
We will simplify the RHS. We the general form of the (r+1) th term in binomial expansion, which is given by ${{T}_{r+1}}={}^{n}{{C}_{r}}{{x}^{r}}$ . We calculate the coefficient of ${{x}^{n}}$ in binomial expansion of ${{\left( 1+x \right)}^{n}}{{\left( x-1 \right)}^{n}}$that will give us the answer.

Complete step-by-step answer:
We know that ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}.........{}^{n}{{C}_{n}}{{b}^{n}}$
We can write the binomial expansion of ${{\left( 1+x \right)}^{n}}$,we will get,
$\Rightarrow {{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}.........{}^{n}{{C}_{n}}{{x}^{n}}$
Similarly, we can write the expansion of ${{\left( x-1 \right)}^{n}}$, we will get,
$\Rightarrow {{\left( x-1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}-{}^{n}{{C}_{1}}{{x}^{n-1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}.........{}^{n}{{C}_{n}}{{\left( -1 \right)}^{n}}$
We will now multiply both the equation, we will get,
$\Rightarrow {{\left( 1+x \right)}^{n}}{{\left( x-1 \right)}^{n}}=\left( {{C}_{0}}^{2}-{{C}_{1}}^{2}+{{C}_{2}}^{2}-{{C}_{3}}^{2}........{{\left( -1 \right)}^{n}}{{C}_{n}}^{2} \right){{x}^{n}}$
From the above equation we can say that ${{C}_{0}}^{2}-{{C}_{1}}^{2}+{{C}_{2}}^{2}-{{C}_{3}}^{2}...........{{\left( -1 \right)}^{n}}{{C}_{n}}^{2}$ is equal to the coefficient of ${{x}^{n}}$ in binomial expansion of ${{\left( 1+x \right)}^{n}}{{\left( x-1 \right)}^{n}}$.
$\Rightarrow {{\left( 1+x \right)}^{n}}{{\left( x-1 \right)}^{n}}={{\left( {{x}^{2}}-1 \right)}^{n}}$
We the general form of the (r+1) th term in binomial expansion, is given by
${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$
We will now write the general term for expansion ${{\left( {{x}^{2}}-1 \right)}^{n}}$
$\Rightarrow {}^{n}{{C}_{r}}{{x}^{2\left( n-r \right)}}{{\left( -1 \right)}^{r}}$
As we can see that for coefficient of ${{x}^{n}}$ ,
$\begin{aligned} & \Rightarrow n=2(n-r) \\\ & \Rightarrow r=\dfrac{n}{2} \\\ \end{aligned}$
Coefficient of can be written as,

& \Rightarrow {}^{n}{{C}_{\dfrac{n}{2}}}{{\left( -1 \right)}^{\dfrac{n}{2}}} \\\ & \Rightarrow {}^{n}{{C}_{\dfrac{n}{2}}}{{\left( -1 \right)}^{\dfrac{n}{2}}} \\\ \end{aligned}$$ Putting n=10, we get, $$\Rightarrow {}^{10}{{C}_{5}}{{\left( -1 \right)}^{5}}$$ Therefore, ${{C}_{0}}^{2}-{{C}_{1}}^{2}+{{C}_{2}}^{2}-{{C}_{3}}^{2}...........{{\left( -1 \right)}^{n}}{{C}_{n}}^{2}={}^{10}{{C}_{5}}{{\left( -1 \right)}^{5}}$ **So, the correct answer is “Option A”.** **Note:** For attending this question we must know the basic concepts of expansion using binomial theorem. We must keep in mind the negative sign that comes with the even terms of expansion of ${{\left( x-1 \right)}^{n}}$. The student must know the general form of the rth term. Don’t get confused while solving ${{\left( 1+x \right)}^{n}}{{\left( x-1 \right)}^{n}}$ simply apply $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ as it is in this form.