1. **Problem Statement:** Use Newton's forward interpolation formula to calculate the values of $s_n$, $2s_n$, $3s_n$, $4s_n$, $5s_n$, and $k s_n$ for $k=5$ given $s_n = n^3 = \frac{n(n+1)}{2}$ and $k s_n = k 1 s_{n+1}^2$, with $k 1 s_n k = 12$ and $s_n = n^3$. 2. **Given:** - $s_n = n^3$ - $k s_n = k 1 s_{n+1}^2$ - $k 1 s_n k = 12$ 3. **Step 1: Calculate $s_n$, $2s_n$, $3s_n$, $4s_n$, $5s_n$, and $k s_n$ for $k=5$** - $s_n = n^3$ - $2s_n = 2 \times n^3 = 2n^3$ - $3s_n = 3 \times n^3 = 3n^3$ - $4s_n = 4 \times n^3 = 4n^3$ - $5s_n = 5 \times n^3 = 5n^3$ - For $k=5$, $k s_n = 5 s_n = 5 n^3$ 4. **Step 2: Taking $n=1$, calculate $s_1$, $2s_1$, $3s_1$, $4s_1$** - $s_1 = 1^3 = 1$ - $2s_1 = 2 \times 1 = 2$ - $3s_1 = 3 \times 1 = 3$ - $4s_1 = 4 \times 1 = 4$ 5. **Step 3: Deduce $s = \sum_{k=0}^4 k s_1$** - $s = 0 \times s_1 + 1 \times s_1 + 2 \times s_1 + 3 \times s_1 + 4 \times s_1$ - $s = (0 + 1 + 2 + 3 + 4) \times 1 = 10$ **Final answer:** - $s_n = n^3$ - $2s_n = 2n^3$ - $3s_n = 3n^3$ - $4s_n = 4n^3$ - $5s_n = 5n^3$ - For $n=1$, $s_1 = 1$, $2s_1 = 2$, $3s_1 = 3$, $4s_1 = 4$ - $s = 10$