Question: $2$. $\sum_{j=2}^{6}(j^3-1)$
**Problem:** Evaluate $\sum_{j=2}^{6}(j^3-1)$.
1. A summation adds the expression for every integer from the lower limit to the upper limit, including both endpoints.
2. Substitute $j=2,3,4,5,6$ into $j^3-1$.
$$\sum_{j=2}^{6}(j^3-1)=(2^3-1)+(3^3-1)+(4^3-1)+(5^3-1)+(6^3-1)$$
3. Evaluate each term.
$$=(8-1)+(27-1)+(64-1)+(125-1)+(216-1)$$
$$=7+26+63+124+215$$
4. Add the terms.
$$7+26+63+124+215=435$$
**Final answer:** $435$.