Question

$\sum _{k=0}^{20}{\left({}^{20}{C}_k\right)}^2$ is equal to

• A. ${}^{41}{C}_{20}$
• B. ${}^{40}{C}_{19}$
• C. ${}^{40}{C}_{21}$
• D. ${}^{40}{C}_{20}$

Answer 1

The correct option is D ${}^{40}{C}_{20}$

We know, $(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x+.....{+}^n{C}_n{x}^n$
Again $(x+1{)}^n{=}^n{C}_0{x}^n{+}^n{C}_1{x}^{n-1}+.....{+}^n{C}_n$
Now, on multiplying both the expressions, we have on the left hand side $(1+x{)}^{2n}$.
On the right hand side we have product of binomial coefficients, and on collecting the required expression, ${(}^n{C}_0{)}^2+{(}^n{C}_1{)}^2+\dots +{(}^n{C}_n{)}^2$, we observe that these are the sum of binomial coefficients which are associated with $x^n$.
Thus, we have
${(}^n{C}_0{)}^2+{(}^n{C}_1{)}^2+\dots +{(}^n{C}_n{)}^2={}^{2n}{C}_n$
$\therefore \sum _{k=0}^{20}{\left({}^{20}{C}_k\right)}^2={(}^{20}{C}_1{)}^2+{(}^{20}{C}_1{)}^2+{(}^{20}{C}_2{)}^2+........+{(}^{20}{C}_{20}{)}^2$
$\therefore \sum _{k=0}^{20}{\left({}^{20}{C}_k\right)}^2={}^{40}{C}_{20}$