To solve the problem of finding the sum ( S = \binom{n}{0} + 3\binom{n}{1} + 5\binom{n}{2} + \dots + (2k+1)\binom{n}{k} + \dots + (2n+1)\binom{n}{n} ), we can break it down into two simpler sums:

Step 1: Split the sum

The general term is ((2k+1)\binom{n}{k}), so:
[ S = \sum{k=0}^n (2k+1)\binom{n}{k} = 2\sum{k=0}^n k\binom{n}{k} + \sum_{k=0}^n \binom{n}{k} ]

Step 2: Compute each part

• Sum of binomial coefficients: (\sum_{k=0}^n \binom{n}{k} = 2^n) (by binomial theorem).
• Sum of (k\binom{n}{k}): Using the identity (k\binom{n}{k} = n\binom{n-1}{k-1}):
  [ \sum{k=0}^n k\binom{n}{k} = n\sum{k=1}^n \binom{n-1}{k-1} = n \cdot 2^{n-1} ]

Step 3: Combine the results

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

Answer: (\boxed{(n+1)2^n}) (or if a specific value is needed, but based on the pattern, the general form is this). For example, if (n) is given (like n=3, it's 32), but the general answer is ((n+1)2^n).

Assuming the problem expects the general formula, the final answer is (\boxed{(n+1)2^n}). If a numerical value is needed (e.g., n=3), it's (\boxed{32}), but likely the general form is intended.

But given common problems, the answer is (\boxed{(n+1)2^n}). However, if the problem had a specific n (like n=5), it would be (5+1)*32=192, but since the original problem's details are a bit unclear, the general formula is the solution.

(\boxed{(n+1)2^n})

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