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Question: The sum of the series \({}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-......

The sum of the series 20C0−20C1+20C2−20C3+....−....+20C10{}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-....+{}^{20}{{C}_{10}} is
A. 1220C10\dfrac{1}{2}{}^{20}{{C}_{10}}
B. 0
C. −20C10-{}^{20}{{C}_{10}}
D. 20C10{}^{20}{{C}_{10}}

Explanation

Solution

To solve this problem, first of all we can start solving this equation from (1+x)20{{(1+x)}^{20}}. When we expand this we will get 20C0+20C1x+20C2x2+20C3x3+....+20C10x20{}^{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 nCr=nCn−r{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}.

Complete step-by-step answer:
First of all, we can consider the equation (1+x)20{{(1+x)}^{20}}. The expansion of (1+x)20{{(1+x)}^{20}} is,
(1+x)20{{(1+x)}^{20}} = 20C0+20C1x+20C2x2+20C3x3+....+20C20x20{}^{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{{(1+(-1))}^{20}} = 20C0+20C1(−1)+20C2(−1)2+20C3(−1)3+....+20C20(−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=20C0−20C1+20C2−20C3+....+20C200={}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....+{}^{20}{{C}_{20}}
We can rewrite this equation using the equation nCr=nCn−r{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}. This, means 20C0=20C20,20C1=20C19.....20C9=20C11{}^{20}{{C}_{0}}={}^{20}{{C}_{20}},{}^{20}{{C}_{1}}={}^{20}{{C}_{19}}.....{}^{20}{{C}_{9}}={}^{20}{{C}_{11}}.
So, the equation becomes,
0=2(20C0−20C1+20C2−20C3+....−20C9)+20C100=2({}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}})+{}^{20}{{C}_{10}}
On further solving, we get,
−20C10=2(20C0−20C1+20C2−20C3+....−20C9)-{}^{20}{{C}_{10}}=2({}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}})
So, we get,
−20C102=20C0−20C1+20C2−20C3+....−20C9\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 20C0−20C1+20C2−20C3+....−....+20C10{}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-....+{}^{20}{{C}_{10}}. So, we can write this equation as,
20C0−20C1+20C2−20C3+....−20C9+20C10{}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}}+{}^{20}{{C}_{10}}= −20C102+20C10\dfrac{-{}^{20}{{C}_{10}}}{2}+{}^{20}{{C}_{10}}
On solving get,
20C0−20C1+20C2−20C3+....−20C9+20C10{}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}}+{}^{20}{{C}_{10}}= 20C102\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{{(1+x)}^{20}}. This is used so that we can easily solve the problem. We know that (1+x)20{{(1+x)}^{20}} = 20C0+20C1x+20C2x2+20C3x3+....+20C20x20{}^{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 nCr=nCn−r{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}. This equation help us to add the common terms in series. 20C10=20C10{}^{20}{{C}_{10}}={}^{20}{{C}_{10}} so we got 2(20C0−20C1+20C2−20C3+....−20C9)+20C102({}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}})+{}^{20}{{C}_{10}}. In the last step, we added 20C10{}^{20}{{C}_{10}} to −20C102\dfrac{-{}^{20}{{C}_{10}}}{2} because −20C102\dfrac{-{}^{20}{{C}_{10}}}{2} is the value of the terms 20C0−20C1+20C2−20C3+....−20C9{}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-{}^{20}{{C}_{9}}. As we are asked to find the value of 20C0−20C1+20C2−20C3+....−....+20C10{}^{20}{{C}_{0}}-{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}-{}^{20}{{C}_{3}}+....-....+{}^{20}{{C}_{10}} we have to add 20C10{}^{20}{{C}_{10}} to −20C102\dfrac{-{}^{20}{{C}_{10}}}{2}.