Question

If \( n \) is a positive integer, the value of:
\[ (2n + 1) \binom{n}{0} + (2n - 1) \binom{n}{1} + (2n - 3) \binom{n}{2} + \dots + 1 \cdot \binom{n}{n} \] is:

• A. \( (n + 1) \cdot 2^n \)
• B. \( 3^n \)
• C. \( f'(2) \text{ where } f(x) = x^{n+1} \)
• D. \( (n + 1) \cdot 2^{n+1} \)

Hint:

For series involving binomial coefficients, split into summations and use binomial properties to simplify. Test alternate representations for verification.

Answer 1

The Correct Option is A, C

1. Step 1: The given series can be expressed as:

\( S = \sum_{k=0}^n \big(2n + 1 - 2k\big) \binom{n}{k}. \)

2. Step 2: Split the summation into two parts:

\( S = (2n + 1) \sum_{k=0}^n \binom{n}{k} - 2 \sum_{k=0}^n k \binom{n}{k}. \)

3. Step 3: Use the binomial summation properties:

• \( \sum_{k=0}^n \binom{n}{k} = 2^n, \)
• \( \sum_{k=0}^n k \binom{n}{k} = n \cdot 2^{n-1}. \)

4. Step 4: Substitute these results:

\( S = (2n + 1) \cdot 2^n - 2 \cdot n \cdot 2^{n-1}. \)

5. Step 5: Simplify:

\( S = (2n + 1) \cdot 2^n - n \cdot 2^n = (n + 1) \cdot 2^n. \)

6. Additionally: Consider \( f(x) = (1 + x)^n (1 - x)^n = x^{n+1} \), then \( f'(x) = (n + 1)x^n. \)

Substituting \( x = 2 \), we also get the same result.

Thus, the correct answers are (A) and (C).