Question

If \((1+x)^n = \sum_{r=0}^n \binom{n}{r} x^r\), then the value of \[ C_0 + (C_0 + C_1) + (C_0 + C_1 + C_2) + \cdots + (C_0 + C_1 + \cdots + C_n) \] is:

• A. \(n 2^{n-1}\)
• B. \(2^n + n\)
• C. \((n + 2) 2^n\)
• D. \((n + 2) 2^{n-1}\)

Hint:

Use combinatorial identities to simplify sums involving binomial coefficients.

Answer 1

The Correct Option is D

Step 1: Express the sum
\[ S = \sum_{k=0}^n \sum_{r=0}^k \binom{n}{r} \] Step 2: Change order of summation
Each \(\binom{n}{r}\) appears in the sums for all \(k \ge r\), so it appears \((n - r + 1)\) times. \[ S = \sum_{r=0}^n (n - r + 1) \binom{n}{r} = \sum_{r=0}^n (n + 1 - r) \binom{n}{r} \] Step 3: Split sum
\[ S = (n+1) \sum_{r=0}^n \binom{n}{r} - \sum_{r=0}^n r \binom{n}{r} \] Step 4: Use identities
\[ \sum_{r=0}^n \binom{n}{r} = 2^n \] \[ \sum_{r=0}^n r \binom{n}{r} = n 2^{n-1} \] Step 5: Final value
\[ S = (n+1) 2^n - n 2^{n-1} = (n+1) 2^n - n 2^{n-1} = 2^{n-1} (2(n+1) - n) = 2^{n-1} (n + 2) \]