Question

$\sum_{k=0}^{20} (^{20}C_k)^2$ is equal to :

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

Hint:

The general Vandermonde's Identity is $\sum_{k=0}^r (^mC_k)(^nC_{r-k}) = ^{m+n}C_r$. Putting $m=n=r=20$ directly gives the answer.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a standard binomial coefficient identity. We can derive it by comparing the coefficient of $x^n$ in the expansion of $(1+x)^n(x+1)^n$.
Step 2: Key Formula or Approach:
Identity: $\sum_{r=0}^n (^nC_r)^2 = ^{2n}C_n$.
Step 3: Detailed Explanation:
We know that $^nC_r = ^nC_{n-r}$.
The sum is $S = \sum_{k=0}^{20} (^{20}C_k) \cdot (^{20}C_{20-k})$.
This sum represents the coefficient of $x^{20}$ in the product of two binomial expansions:
\[ (1+x)^{20} \cdot (1+x)^{20} = (1+x)^{40} \]
Coefficient of $x^{20}$ in $(1+x)^{40}$ is $^{40}C_{20}$.
Therefore, $\sum_{k=0}^{20} (^{20}C_k)^2 = ^{40}C_{20}$.
Step 4: Final Answer:
The sum is $^{40}C_{20}$.