Question

If $ \sum_{k=0}^{n+1} C_k^n = 512 $, find $ \sum_{k=0}^{n} C_k^n $.

• A. \( 512 \)
• B. \( 256 \)
• C. \( 1024 \)
• D. \( 1023 \)

Hint:

For binomial coefficients, remember that the sum \( \sum_{k=0}^{n} C_k^n \) always equals \( 2^n \), and consider the properties of binomial coefficients when solving such problems.

Answer 1

The Correct Option is A

Let $ \sum_{k=0}^{n+1} C_k^n = \sum_{k=0}^{n} \binom{n}{k} + \binom{n}{n+1}$. 

We understand that when n is a positive integer, $\binom{n}{k}= 0$ if $k > n$ and also if $k <0$. 

Thus $\binom{n}{n+1} = 0$, leading to $\sum_{k=0}^{n} \binom{n}{k}=2^n$ and $\sum_{k=0}^{n+1} C_k^n =2^n$.

Thus $2^n = 512=2^9$ so n=9. 

As before the summation from k=0 to n is also 512.

Final Answer: The final answer is $\boxed{512}$