Question

The sum of the series ${}^{2020}{C_0} - {}^{2020}{C_1} + {}^{2020}{C_2} - {}^{2020}{C_3} + ... + {}^{2020}{C_{1010}}$ is-
A.$\dfrac{1}{2}{}^{2020}{C_{1010}}$
B.${}^{2020}{C_{1010}}$
C.Zero
D.None of these

Answer 1

Hint: We know that the expansion of ${\left( {1 + x} \right)^n}$ is given as-
$ \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$ and n=$2020$and then solve the expansion obtained. Then we know that $ - {1^n} = 1$ if n is an even number and $ - {1^n} = - 1$ if n is an odd number. Use this to simplify further. Then use the formula of combination given as-${}^n{C_r} = {}^n{C_{n - r}}$
Then we will get two same terms except ${}^{2020}{C_{1010}}$ 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-${}^{2020}{C_0} - {}^{2020}{C_1} + {}^{2020}{C_2} - {}^{2020}{C_3} + ... + {}^{2020}{C_{1010}}$
We know that-
$ \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$ and n=$2020$ , we get-
$ \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-
$ \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 ${1^n} = 1$ where n is a positive integer.
Then, on further simplifying, we get-
$ \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 $ - {1^n} = 1$ if n is an even number and $ - {1^n} = - 1$ if n is an odd number. Then on applying this, we get-
$ \Rightarrow 0 = {}^{2020}{C_0} - {}^{2020}{C_1} + ... + {}^{2020}{C_{2020}}$
Now we know that ${}^n{C_r} = {}^n{C_{n - r}}$
Then we can write-
\[ \Rightarrow {}^{2020}{C_0} = {}^{2020}{C_{2020 - 0}} = {}^{2020}{C_{2020}}\]
Similarly we can also write-
$ \Rightarrow $ \[{}^{2020}{C_1} = {}^{2020}{C_{2020 - 1}} = {}^{2020}{C_{2019}}\]
\[ \Rightarrow {}^{2020}{C_2} = {}^{2020}{C_{2020 - 2}} = {}^{2020}{C_{2018}}\]
--- “--- “---- “-----
$ \Rightarrow {}^{2020}{C_{1009}} = {}^{2020}{C_{2020 - 1009}} = {}^{2020}{C_{1011}}$
On substituting these values in the equation we get-
\[ \Rightarrow 0 = {}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}} + {}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}}\]
Here we can see that except ${}^{2020}{C_{1010}}$ every combination comes two times so on adding this term to the both side we get-
\[ \Rightarrow {}^{2020}{C_{1010}} = {}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}} + {}^{2020}{C_{1010}} + {}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}}\]
Now on adding the given terms we get-
\[ \Rightarrow {}^{2020}{C_{1010}} = 2{}^{2020}{C_0} - 2{}^{2020}{C_1} + ... - 2{}^{2020}{C_{1009}} + 2{}^{2020}{C_{1010}}\]
Now on taking $2$ common we get-
\[ \Rightarrow {}^{2020}{C_{1010}} = 2\left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}}} \right]\]
On transferring the $2$ from the left side to the right side, we get-
\[ \Rightarrow \dfrac{{{}^{2020}{C_{1010}}}}{2} = \left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}}} \right]\]
We can also write it as-
\[ \Rightarrow \left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}}} \right] = \dfrac{1}{2}{}^{2020}{C_{1010}}\]
The correct answer is option A.


Note: Here, the student may go wrong if they do not add the term ${}^{2020}{C_{1010}}$ both side because then we will get-
\[ \Rightarrow - {}^{2020}{C_{1010}} = 2{}^{2020}{C_0} - 2{}^{2020}{C_1} + ... - 2{}^{2020}{C_{1009}}\]
On solving this we will get the sum of the series-
\[ \Rightarrow \dfrac{{ - {}^{2020}{C_{1010}}}}{2} = \left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}}} \right]\]
But we have to find the sum of the series-
$ \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-
\[ \Rightarrow {}^{2020}{C_{1010}} + \dfrac{{ - {}^{2020}{C_{1010}}}}{2} = \left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}}} \right]\]
On solving this we will get-
\[ \Rightarrow \left[ {{}^{2020}{C_0} - {}^{2020}{C_1} + ... - {}^{2020}{C_{1009}} + {}^{2020}{C_{1010}}} \right] = \dfrac{1}{2}{}^{2020}{C_{1010}}\]