Question

Find the value of given binomial expansion ${}^{100}{{C}_{0}}{}^{200}{{C}_{100}}-{}^{100}{{C}_{1}}{}^{199}{{C}_{100}}+{}^{100}{{C}_{2}}{}^{198}{{C}_{100}}-......+{}^{100}{{C}_{100}}{}^{100}{{C}_{100}}$ .
(a) 1
(b) -1
(c) 0
(d) 2

Answer 1

Hint : Here we observe that ${}^{200}{{C}_{100}},{}^{199}{{C}_{100}},{}^{198}{{C}_{100}},......,{}^{100}{{C}_{100}}$ are the coefficient of ${{x}^{100}}$ in binomial expansion of ${{\left( 1+x \right)}^{200}},{{\left( 1+x \right)}^{199}},{{\left( 1+x \right)}^{198}},......,{{\left( 1+x \right)}^{100}}$ . We can then see that the remaining expansion resembles the expansion of ${{\left( y-1 \right)}^{n}}$ . Using all these facts we can get the value of required expansion.

Complete step-by-step answer :
Given that we need to find the value of given binomial expansion ${}^{100}{{C}_{0}}{}^{200}{{C}_{100}}-{}^{100}{{C}_{1}}{}^{199}{{C}_{100}}+{}^{100}{{C}_{2}}{}^{198}{{C}_{100}}-......+{}^{100}{{C}_{100}}{}^{100}{{C}_{100}}---(1)$
We know that coefficient of ${{x}^{r}}$ in ${{(1+x)}^{n}}$ is ${}^{n}{{C}_{r}}$ .
By using this fact, we can observe that ${}^{200}{{C}_{100}}$ is co-efficient for ${{x}^{100}}$ in binomial expansion of ${{\left( 1+x \right)}^{200}}$ .
We can also observe that ${}^{199}{{C}_{100}},{}^{198}{{C}_{100}},......,{}^{100}{{C}_{100}}$ are the coefficient of ${{x}^{100}}$ in binomial expansion of ${{\left( 1+x \right)}^{199}},{{\left( 1+x \right)}^{198}},......,{{\left( 1+x \right)}^{100}}$ .
So, we can write ${}^{100}{{C}_{0}}{}^{200}{{C}_{100}}-{}^{100}{{C}_{1}}{}^{199}{{C}_{100}}+{}^{100}{{C}_{2}}{}^{198}{{C}_{100}}-......+{}^{100}{{C}_{100}}{}^{100}{{C}_{100}}$ as coefficient of ${{x}^{100}}$ in binomial expansion $\left[ {}^{100}{{C}_{0}}{{\left( 1+x \right)}^{200}}-{}^{100}{{C}_{1}}{{\left( 1+x \right)}^{199}}+{}^{100}{{C}_{2}}{{\left( 1+x \right)}^{198}}-......+{}^{100}{{C}_{100}}{{\left( 1+x \right)}^{100}} \right]$.
We can take ${{\left( 1+x \right)}^{100}}$ common from all the terms of binomial expansion.
So, we can write it as coefficient of ${{x}^{100}}$ in binomial expansion ${{\left( 1+x \right)}^{100}}\times \left[ {}^{100}{{C}_{0}}{{\left( 1+x \right)}^{100}}-{}^{100}{{C}_{1}}{{\left( 1+x \right)}^{99}}+{}^{100}{{C}_{2}}{{\left( 1+x \right)}^{98}}-......+{}^{100}{{C}_{100}}.1 \right]---(2)$.
We know that expansion of ${{\left( y-1 \right)}^{n}}$ is ${}^{n}{{C}_{0}}{{y}^{n}}-{}^{n}{{C}_{1}}{{y}^{n-1}}+{}^{n}{{C}_{2}}{{y}^{n-2}}-......+{}^{n}{{C}_{n}}{{\left( -1 \right)}^{n}}{{y}^{n}}$ .
We can now see that the expansion inside square brackets of equation (2) resembles the expansion of ${{\left( y-1 \right)}^{n}}$ .
Here we get $y=(1+x)$ and $n=100$ .
Using all these equation (1) can be written as coefficient of ${{x}^{100}}$ in ${{\left( 1+x \right)}^{100}}\times {{\left( 1+x-1 \right)}^{100}}$ .
Now we can write equation (1) as coefficient of ${{x}^{100}}$ in ${{\left( 1+x \right)}^{100}}\times {{\left( x \right)}^{100}}$ .
We know that expansion of ${{(1+x)}^{n}}$ is ${}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}.x+{}^{n}{{C}_{2}}.{{x}^{2}}+......+{}^{n}{{C}_{n}}.{{x}^{n}}$ .
Now we apply this expansion for ${{\left( 1+x \right)}^{100}}$ . Here we take n = 100.
Equation (1) can be written as coefficient of ${{x}^{100}}$ in $\left( {}^{100}{{C}_{0}}+{}^{100}{{C}_{1}}.x+{}^{100}{{C}_{2}}.{{x}^{2}}+......+{}^{100}{{C}_{100}}.{{x}^{100}} \right)\times {{x}^{100}}$ .
Equation (1) can now be written as a coefficient of ${{x}^{100}}$ in $\left( {}^{100}{{C}_{0}}.{{x}^{100}}+{}^{100}{{C}_{1}}.{{x}^{101}}+{}^{100}{{C}_{2}}.{{x}^{102}}+......+{}^{100}{{C}_{100}}.{{x}^{200}} \right)$ .
Since, we need a value of coefficient of ${{x}^{100}}$ . We get ${}^{100}{{C}_{0}}$ by observing from the expansion.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!.(n-r)!}$ , where ‘!’ represents factorial and the value of n! is $n!=n\times (n-1)\times (n-2)\times ......\times 1$ and value of 0! = 1.
So, ${}^{100}{{C}_{0}}=\dfrac{100!}{0!(100-0)!}$.
${}^{100}{{C}_{0}}=\dfrac{100!}{0!100!}$.
${}^{100}{{C}_{0}}=\dfrac{100!}{1.100!}$
${}^{100}{{C}_{0}}=1$ .
∴ The value of ${}^{100}{{C}_{0}}{}^{200}{{C}_{100}}-{}^{100}{{C}_{1}}{}^{199}{{C}_{100}}+{}^{100}{{C}_{2}}{}^{198}{{C}_{100}}-......+{}^{100}{{C}_{100}}{}^{100}{{C}_{100}}$ is 1.
So, the correct answer is “Option A”.

Note : This type of problem looks difficult to see, but if you get the logic it will be easy to solve. Whenever we get problems involving multiplication of two binomial coefficients, we need to check whether one of them coincides with one of coefficients of a binomial expansion. After checking such coefficients we move towards answers in an efficient way.