Question

The sum of the series ${}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-....+{}^{20}{{C}_{10}}$ is
A. $\dfrac{1}{2}{}^{20}{{C}_{10}}$
B. 0
C. $-{}^{20}{{C}_{10}}$
D. ${}^{20}{{C}_{10}}$

Answer 1

Hint: To solve this problem, first of all we can start solving this equation from ${{(1+x)}^{20}}$. When we expand this we will get ${}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}x+{}^{20}{{C}_{2}}{{x}^{2}}+{}^{20}{{C}_{3}}{{x}^{3}}+....+{}^{20}{{C}_{10}}{{x}^{20}}$. After this, we can replace the value of x with -1. Then expand the whole equation and we will get alternative positive and negative terms. We can use the equation such as ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$.

Complete step-by-step answer:
First of all, we can consider the equation ${{(1+x)}^{20}}$. The expansion of ${{(1+x)}^{20}}$ is,
${{(1+x)}^{20}}$ = ${}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}x+{}^{20}{{C}_{2}}{{x}^{2}}+{}^{20}{{C}_{3}}{{x}^{3}}+....+{}^{20}{{C}_{20}}{{x}^{20}}$
Now, we can put x = -1. So, we get,
${{(1+(-1))}^{20}}$ = ${}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}(-1)+{}^{20}{{C}_{2}}{{(-1)}^{2}}+{}^{20}{{C}_{3}}{{(-1)}^{3}}+....+{}^{20}{{C}_{20}}{{(-1)}^{20}}$
On expanding the above equation we get,
$0={}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....+{}^{20}{{C}_{20}}$
We can rewrite this equation using the equation ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$. This, means ${}^{20}{{C}_{0}}={}^{20}{{C}_{20}},{}^{20}{{C}_{1}}={}^{20}{{C}_{19}}.....{}^{20}{{C}_{9}}={}^{20}{{C}_{11}}$.
So, the equation becomes,
$0=2({}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}})+{}^{20}{{C}_{10}}$
On further solving, we get,
$-{}^{20}{{C}_{10}}=2({}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}})$
So, we get,
$\dfrac{-{}^{20}{{C}_{10}}}{2}={}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}}$
Now, we are asked to find the value of ${}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-....+{}^{20}{{C}_{10}}$. So, we can write this equation as,
${}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}}+{}^{20}{{C}_{10}}$= $\dfrac{-{}^{20}{{C}_{10}}}{2}+{}^{20}{{C}_{10}}$
On solving get,
${}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}}+{}^{20}{{C}_{10}}$= $\dfrac{{}^{20}{{C}_{10}}}{2}$

So, the correct answer is “Option A”.

Note: In this problem, we have started solving using the ${{(1+x)}^{20}}$. This is used so that we can easily solve the problem. We know that ${{(1+x)}^{20}}$ = ${}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}x+{}^{20}{{C}_{2}}{{x}^{2}}+{}^{20}{{C}_{3}}{{x}^{3}}+....+{}^{20}{{C}_{20}}{{x}^{20}}$. Here, x is replaced by -1 because in the question the terms are in alternate positive and negative terms. We have used few general equation of the combination and that is ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$. This equation help us to add the common terms in series. ${}^{20}{{C}_{10}}={}^{20}{{C}_{10}}$ so we got $2({}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}})+{}^{20}{{C}_{10}}$. In the last step, we added ${}^{20}{{C}_{10}}$ to $\dfrac{-{}^{20}{{C}_{10}}}{2}$ because $\dfrac{-{}^{20}{{C}_{10}}}{2}$ is the value of the terms ${}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}}$. As we are asked to find the value of ${}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-....+{}^{20}{{C}_{10}}$ we have to add ${}^{20}{{C}_{10}}$ to $\dfrac{-{}^{20}{{C}_{10}}}{2}$.