1. The problem involves evaluating the sum $$\sum_{k=1}^{n+1} a + k C_{k-1} d$$. 2. First, clarify the expression: it appears to be a sum from $k=1$ to $n+1$ of the terms $a + k \cdot C_{k-1} \cdot d$, where $C_{k-1}$ likely denotes a binomial coefficient or a constant term. 3. Assuming $C_{k-1}$ is a binomial coefficient, for example $C_{k-1} = \binom{n}{k-1}$, the sum becomes $$\sum_{k=1}^{n+1} a + k \binom{n}{k-1} d$$. 4. We can separate the sum into two sums: $$\sum_{k=1}^{n+1} a + \sum_{k=1}^{n+1} k \binom{n}{k-1} d$$. 5. The first sum is $$a \times (n+1)$$ because $a$ is constant and there are $n+1$ terms. 6. For the second sum, change the index by letting $j = k-1$, so when $k=1$, $j=0$ and when $k=n+1$, $j=n$: $$\sum_{j=0}^n (j+1) \binom{n}{j} d$$. 7. This sum can be split as: $$d \sum_{j=0}^n j \binom{n}{j} + d \sum_{j=0}^n \binom{n}{j}$$. 8. Recall the identities: $$\sum_{j=0}^n \binom{n}{j} = 2^n$$ $$\sum_{j=0}^n j \binom{n}{j} = n 2^{n-1}$$ 9. Substitute these into the sum: $$d (n 2^{n-1} + 2^n) = d 2^{n-1} (n + 2)$$. 10. Therefore, the total sum is: $$a (n+1) + d 2^{n-1} (n + 2)$$. Final answer: $$\sum_{k=1}^{n+1} a + k C_{k-1} d = a (n+1) + d 2^{n-1} (n + 2)$$