1. **State the problem:** We need to find the sum of the series given by $$\sum_{k=1}^{n+1} (a + C (k-1) d)$$ where $a$, $C$, and $d$ are constants. 2. **Understand the series:** This is an arithmetic series because the term inside the summation is linear in $k$: $$a + C (k-1) d$$. 3. **Rewrite the summation:** $$\sum_{k=1}^{n+1} (a + C (k-1) d) = \sum_{k=1}^{n+1} a + \sum_{k=1}^{n+1} C (k-1) d$$ 4. **Separate the sums:** $$= (n+1) a + C d \sum_{k=1}^{n+1} (k-1)$$ 5. **Sum of integers:** Recall the formula for the sum of the first $m$ integers: $$\sum_{k=0}^{m} k = \frac{m(m+1)}{2}$$ Here, $m = n$ because $k-1$ goes from $0$ to $n$ as $k$ goes from $1$ to $n+1$. 6. **Calculate the sum:** $$\sum_{k=1}^{n+1} (k-1) = \sum_{j=0}^{n} j = \frac{n(n+1)}{2}$$ 7. **Substitute back:** $$S = (n+1) a + C d \frac{n(n+1)}{2}$$ 8. **Final answer:** $$\boxed{S = (n+1) a + \frac{C d n (n+1)}{2}}$$ This formula gives the sum of the series for any integer $n \geq 0$.