1. **Problem statement:** We need to simplify and calculate the combined sum: $$\left( \sum_{t=-2}^5 (3t - 2) \right) - \left( \sum_{t=1}^6 (2t + 3) \right) + \left( \sum_{t=-1}^3 (1 - t) \right)$$ 2. **Goal:** Rewrite the sums under a single summation symbol by shifting indices and splitting sums if necessary, then compute the total. 3. **Step 1: Write each sum explicitly and find their ranges.** - First sum: $t$ from $-2$ to $5$ of $3t - 2$ - Second sum: $t$ from $1$ to $6$ of $2t + 3$ - Third sum: $t$ from $-1$ to $3$ of $1 - t$ 4. **Step 2: Express the entire expression as a single sum by adjusting indices.** The combined range of $t$ values is from $-2$ (lowest) to $6$ (highest). Define a new sum from $t = -2$ to $6$ and express each original sum in terms of this range, using zero where terms do not exist. 5. **Step 3: Write each sum as a sum over $t = -2$ to $6$ with zero terms outside original ranges:** $$\sum_{t=-2}^6 \Big[ \underbrace{(3t - 2) \cdot I_{[-2,5]}(t)}_{\text{first sum}} - \underbrace{(2t + 3) \cdot I_{[1,6]}(t)}_{\text{second sum}} + \underbrace{(1 - t) \cdot I_{[-1,3]}(t)}_{\text{third sum}} \Big]$$ where $I_{[a,b]}(t)$ is the indicator function equal to 1 if $t$ in $[a,b]$, else 0. 6. **Step 4: Write out the combined summand for each $t$ from $-2$ to $6$:** - For $t = -2$ to $5$, first sum term is $3t - 2$, else 0. - For $t = 1$ to $6$, second sum term is $2t + 3$, else 0. - For $t = -1$ to $3$, third sum term is $1 - t$, else 0. 7. **Step 5: Calculate the sum term by term:** Calculate each sum separately first: - First sum: $$\sum_{t=-2}^5 (3t - 2) = 3 \sum_{t=-2}^5 t - 2 \sum_{t=-2}^5 1$$ Number of terms: $5 - (-2) + 1 = 8$ Sum of $t$ from $-2$ to $5$: $$\sum_{t=-2}^5 t = \sum_{t=0}^5 t + \sum_{t=-2}^{-1} t = (0+1+2+3+4+5) + (-2 + -1) = 15 - 3 = 12$$ Sum of 1 over 8 terms is 8. So first sum: $$3 \times 12 - 2 \times 8 = 36 - 16 = 20$$ - Second sum: $$\sum_{t=1}^6 (2t + 3) = 2 \sum_{t=1}^6 t + 3 \sum_{t=1}^6 1$$ Number of terms: $6 - 1 + 1 = 6$ Sum of $t$ from 1 to 6: $$1 + 2 + 3 + 4 + 5 + 6 = 21$$ Sum of 1 over 6 terms is 6. So second sum: $$2 \times 21 + 3 \times 6 = 42 + 18 = 60$$ - Third sum: $$\sum_{t=-1}^3 (1 - t) = \sum_{t=-1}^3 1 - \sum_{t=-1}^3 t$$ Number of terms: $3 - (-1) + 1 = 5$ Sum of 1 over 5 terms is 5. Sum of $t$ from $-1$ to $3$: $$-1 + 0 + 1 + 2 + 3 = 5$$ So third sum: $$5 - 5 = 0$$ 8. **Step 6: Combine all sums:** $$20 - 60 + 0 = -40$$ **Final answer:** $$\boxed{-40}$$