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Question: The sum of the series \({}^{2020}{C_0} - {}^{2020}{C_1} + {}^{2020}{C_2} - {}^{2020}{C_3} + ... + {}...

The sum of the series 2020C0−2020C1+2020C2−2020C3+...+2020C1010{}^{2020}{C_0} - {}^{2020}{C_1} + {}^{2020}{C_2} - {}^{2020}{C_3} + ... + {}^{2020}{C_{1010}} is-
A.122020C1010\dfrac{1}{2}{}^{2020}{C_{1010}}
B.2020C1010{}^{2020}{C_{1010}}
C.Zero
D.None of these

Explanation

Solution

We know that the expansion of (1+x)n{\left( {1 + x} \right)^n} is given as-
⇒(1+x)n=nC0(1)nx0+nC1(1)n−1x1+...+nCn(1)n−nxn\Rightarrow {\left( {1 + x} \right)^n} = {}^n{C_0}{\left( 1 \right)^n}{x^0} + {}^n{C_1}{\left( 1 \right)^{n - 1}}{x^1} + ... + {}^n{C_n}{\left( 1 \right)^{n - n}}{x^n}
Here put x=−1 - 1 and n=20202020and then solve the expansion obtained. Then we know that −1n=1 - {1^n} = 1 if n is an even number and −1n=−1 - {1^n} = - 1 if n is an odd number. Use this to simplify further. Then use the formula of combination given as-nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}}
Then we will get two same terms except 2020C1010{}^{2020}{C_{1010}} so add this term to both sides in the equation we obtained and solve to get the answer.

Complete step-by-step answer:
We have to find the sum of the give series-2020C0−2020C1+2020C2−2020C3+...+2020C1010{}^{2020}{C_0} - {}^{2020}{C_1} + {}^{2020}{C_2} - {}^{2020}{C_3} + ... + {}^{2020}{C_{1010}}
We know that-
⇒(1+x)n=nC0(1)nx0+nC1(1)n−1x1+...+nCn(1)n−nxn\Rightarrow {\left( {1 + x} \right)^n} = {}^n{C_0}{\left( 1 \right)^n}{x^0} + {}^n{C_1}{\left( 1 \right)^{n - 1}}{x^1} + ... + {}^n{C_n}{\left( 1 \right)^{n - n}}{x^n}
Then on putting x=−1 - 1 and n=20202020 , we get-
⇒(1+(−1))2020=2020C0(1)2020(−1)0+2020C1(1)2020−1(−1)1+...+2020C2020(1)2020−2020(−1)2020\Rightarrow {\left( {1 + \left( { - 1} \right)} \right)^{2020}} = {}^{2020}{C_0}{\left( 1 \right)^{2020}}{\left( { - 1} \right)^0} + {}^{2020}{C_1}{\left( 1 \right)^{2020 - 1}}{\left( { - 1} \right)^1} + ... + {}^{2020}{C_{2020}}{\left( 1 \right)^{2020 - 2020}}{\left( { - 1} \right)^{2020}}
On simplifying we get-
⇒(0)2020=2020C0(1)2020(−1)0+2020C1(1)2019(−1)1+...+2020C2020(1)0(−1)2020\Rightarrow {\left( 0 \right)^{2020}} = {}^{2020}{C_0}{\left( 1 \right)^{2020}}{\left( { - 1} \right)^0} + {}^{2020}{C_1}{\left( 1 \right)^{2019}}{\left( { - 1} \right)^1} + ... + {}^{2020}{C_{2020}}{\left( 1 \right)^0}{\left( { - 1} \right)^{2020}}
Now we know that 1n=1{1^n} = 1 where n is a positive integer.
Then, on further simplifying, we get-
⇒0=2020C0(−1)0+2020C1(−1)1+...+2020C2020(−1)2020\Rightarrow 0 = {}^{2020}{C_0}{\left( { - 1} \right)^0} + {}^{2020}{C_1}{\left( { - 1} \right)^1} + ... + {}^{2020}{C_{2020}}{\left( { - 1} \right)^{2020}}
Now we know that −1n=1 - {1^n} = 1 if n is an even number and −1n=−1 - {1^n} = - 1 if n is an odd number. Then on applying this, we get-
⇒0=2020C0−2020C1+...+2020C2020\Rightarrow 0 = {}^{2020}{C_0} - {}^{2020}{C_1} + ... + {}^{2020}{C_{2020}}
Now we know that nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}}
Then we can write-
⇒2020C0=2020C2020−0=2020C2020\Rightarrow {}^{2020}{C_0} = {}^{2020}{C_{2020 - 0}} = {}^{2020}{C_{2020}}
Similarly we can also write-
⇒\Rightarrow 2020C1=2020C2020−1=2020C2019{}^{2020}{C_1} = {}^{2020}{C_{2020 - 1}} = {}^{2020}{C_{2019}}
⇒2020C2=2020C2020−2=2020C2018\Rightarrow {}^{2020}{C_2} = {}^{2020}{C_{2020 - 2}} = {}^{2020}{C_{2018}}
--- “--- “---- “-----
⇒2020C1009=2020C2020−1009=2020C1011\Rightarrow {}^{2020}{C_{1009}} = {}^{2020}{C_{2020 - 1009}} = {}^{2020}{C_{1011}}
On substituting these values in the equation we get-
⇒0=2020C0−2020C1+...−2020C1009+2020C1010+2020C0−2020C1+...−2020C1009\Rightarrow 0 = {}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}} + {}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}}
Here we can see that except 2020C1010{}^{2020}{C_{1010}} every combination comes two times so on adding this term to the both side we get-
⇒2020C1010=2020C0−2020C1+...−2020C1009+2020C1010+2020C1010+2020C0−2020C1+...−2020C1009\Rightarrow {}^{2020}{C_{1010}} = {}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}} + {}^{2020}{C_{1010}} + {}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}}
Now on adding the given terms we get-
⇒2020C1010=22020C0−22020C1+...−22020C1009+22020C1010\Rightarrow {}^{2020}{C_{1010}} = 2{}^{2020}{C_0} - 2{}^{2020}{C_1} + ... - 2{}^{2020}{C_{1009}} + 2{}^{2020}{C_{1010}}
Now on taking 22 common we get-
⇒2020C1010=2[2020C0−2020C1+...−2020C1009+2020C1010]\Rightarrow {}^{2020}{C_{1010}} = 2\left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}}} \right]
On transferring the 22 from the left side to the right side, we get-
⇒2020C10102=[2020C0−2020C1+...−2020C1009+2020C1010]\Rightarrow \dfrac{{{}^{2020}{C_{1010}}}}{2} = \left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}}} \right]
We can also write it as-
⇒[2020C0−2020C1+...−2020C1009+2020C1010]=122020C1010\Rightarrow \left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}}} \right] = \dfrac{1}{2}{}^{2020}{C_{1010}}
The correct answer is option A.

Note: Here, the student may go wrong if they do not add the term 2020C1010{}^{2020}{C_{1010}} both side because then we will get-
⇒−2020C1010=22020C0−22020C1+...−22020C1009\Rightarrow - {}^{2020}{C_{1010}} = 2{}^{2020}{C_0} - 2{}^{2020}{C_1} + ... - 2{}^{2020}{C_{1009}}
On solving this we will get the sum of the series-
⇒−2020C10102=[2020C0−2020C1+...−2020C1009]\Rightarrow \dfrac{{ - {}^{2020}{C_{1010}}}}{2} = \left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}}} \right]
But we have to find the sum of the series-
⇒2020C0−2020C1+2020C2−2020C3+...+2020C1010\Rightarrow {}^{2020}{C_0} - {}^{2020}{C_1} + {}^{2020}{C_2} - {}^{2020}{C_3} + ... + {}^{2020}{C_{1010}}
So this will not give as the answer to our question.
So to get the answer, we will have to add this term both side in the last step-
⇒2020C1010+−2020C10102=[2020C0−2020C1+...−2020C1009+2020C1010]\Rightarrow {}^{2020}{C_{1010}} + \dfrac{{ - {}^{2020}{C_{1010}}}}{2} = \left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}}} \right]
On solving this we will get-
⇒[2020C0−2020C1+...−2020C1009+2020C1010]=122020C1010\Rightarrow \left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}}} \right] = \dfrac{1}{2}{}^{2020}{C_{1010}}