Sum Cubes Minus One C68354

1. **State the problem:** We need to evaluate the sum $$\sum_{j=2}^{6} (j^3 - 1)$$ which means adding the values of $j^3 - 1$ for $j=2,3,4,5,6$. 2. **Formula and rules:** The summation notation $$\sum_{j=a}^{b} f(j)$$ means to calculate $f(j)$ for each integer $j$ from $a$ to $b$ and then add all those values. 3. **Calculate each term:** - For $j=2$: $2^3 - 1 = 8 - 1 = 7$ - For $j=3$: $3^3 - 1 = 27 - 1 = 26$ - For $j=4$: $4^3 - 1 = 64 - 1 = 63$ - For $j=5$: $5^3 - 1 = 125 - 1 = 124$ - For $j=6$: $6^3 - 1 = 216 - 1 = 215$ 4. **Sum all terms:** $$7 + 26 + 63 + 124 + 215 = 435$$ 5. **Final answer:** $$\sum_{j=2}^{6} (j^3 - 1) = 435$$