Finding a Partial Sum In Exercises 59-64, find the partial sum. 59. $$ \sum_{n=1}^{50} n $$ 61. $$ \sum_{n=11}^{30} n-\sum_{n=1}^{100} n n $$ 63. $$ \sum_{n=1}^{500}(n+8) $$ 62. $$ \sum_{n=51}^{100} n-\sum_{n=1}^{50} n $$ 64. $$ \sum_{n=1}^{250}(1000-n) $$

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**59.** We wish to evaluate  
$$
\sum_{n=1}^{50} n.
$$  
Recall the formula for the sum of the first $$ N $$ positive integers:  
$$
\sum_{n=1}^{N} n = \frac{N(N+1)}{2}.
$$  
For $$ N=50 $$:  
$$
\sum_{n=1}^{50} n = \frac{50\cdot 51}{2} = \frac{2550}{2} = 1275.
$$

---

**61.** We wish to evaluate  
$$
\sum_{n=11}^{30} n - \sum_{n=1}^{100} n.
$$

_First term: $$ \sum_{n=11}^{30} n $$._

The sum from 1 to 30 is  
$$
\sum_{n=1}^{30} n = \frac{30\cdot 31}{2} = 465.
$$  
The sum from 1 to 10 is  
$$
\sum_{n=1}^{10} n = \frac{10\cdot 11}{2} = 55.
$$  
Thus, the sum from 11 to 30 is  
$$
\sum_{n=11}^{30} n = \sum_{n=1}^{30} n - \sum_{n=1}^{10} n = 465 - 55 = 410.
$$

_Second term: $$ \sum_{n=1}^{100} n $$._

Using the formula with $$ N=100 $$:  
$$
\sum_{n=1}^{100} n = \frac{100\cdot 101}{2} = 5050.
$$

_Therefore, the desired expression is:_
$$
410 - 5050 = -4640.
$$

---

**62.** We wish to evaluate  
$$
\sum_{n=51}^{100} n - \sum_{n=1}^{50} n.
$$

_First, compute $$ \sum_{n=51}^{100} n $$._

Rather than summing directly, notice that  
$$
\sum_{n=1}^{100} n = \frac{100\cdot 101}{2} = 5050
$$  
and  
$$
\sum_{n=1}^{50} n = \frac{50\cdot 51}{2} = 1275.
$$  
Thus,  
$$
\sum_{n=51}^{100} n = \sum_{n=1}^{100} n - \sum_{n=1}^{50} n = 5050 - 1275 = 3775.
$$

_Now subtract the second sum:_
$$
3775 - 1275 = 2500.
$$

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**63.** We wish to evaluate  
$$
\sum_{n=1}^{500} (n+8).
$$

Break the sum into two parts:  
$$
\sum_{n=1}^{500}(n+8) = \sum_{n=1}^{500} n + \sum_{n=1}^{500} 8.
$$

_First part:_
$$
\sum_{n=1}^{500} n = \frac{500\cdot 501}{2} = 125250.
$$

_Second part:_ There are 500 terms, each contributing 8, so  
$$
\sum_{n=1}^{500} 8 = 8\cdot 500 = 4000.
$$

_Thus, the total is:_
$$
125250 + 4000 = 129250.
$$

---

**64.** We wish to evaluate  
$$
\sum_{n=1}^{250}(1000-n).
$$

Again, separate into two sums:  
$$
\sum_{n=1}^{250}(1000-n) = \sum_{n=1}^{250} 1000 - \sum_{n=1}^{250} n.
$$

_First part:_  
$$
\sum_{n=1}^{250} 1000 = 1000\cdot 250 = 250000.
$$

_Second part:_ Use the formula for the sum of the first 250 integers:  
$$
\sum_{n=1}^{250} n = \frac{250\cdot 251}{2}.
$$  
Compute:
$$
250\cdot 251 = 62750,\quad \text{so}\quad \frac{62750}{2} = 31375.
$$

_Therefore, the sum is:_
$$
250000 - 31375 = 218625.
$$

---

Final Answers:

- 59. $$1275$$  
- 61. $$-4640$$  
- 62. $$2500$$  
- 63. $$129250$$  
- 64. $$218625$$
- 59. 1275 - 61. -4640 - 62. 2500 - 63. 129250 - 64. 218625

Finding a Partial Sum
In Exercises 59-64, find the
partial sum.
59.
61.
63.
62.
64.

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Finding a Partial Sum In Exercises 59-64, find the partial sum. 59. 61. 63. 62. 64.