Chapter 4: Q. 35 (page 326)

Find a formula for each of the sums in Exercises 35, and then use these formulas to calculate each sum for $n = 100, n = 500$ and $n = 1000$ .
$\sum_{k = 1}^{n} (3 - k)$

Short Answer

Expert verified

The sum is $\begin{array}{rcl} \frac{5 n - n^{2}}{2} \end{array}$ .

The sum when $n = 100$ is $- 4750$ .

The sum when $n = 500$ is $- 123750$ .

The sum when $n = 1000$ is $- 497500$ .

Step by step solution

Given information

The given summation is $\sum_{k = 1}^{n} (3 - k)$ .

Determine the formula for the given summation.

The sum can be written as:

$\begin{array}{rcl} \sum_{k = 1}^{n} (3 - k) & = & 3 \sum_{k = 1}^{n} 1 - \sum_{k = 1}^{n} k \\ = & 3 n - \frac{n (n + 1)}{2} [\sum_{k = 1}^{n} k = \frac{n (n + 1)}{2} and \sum_{k = 1}^{n} 1 = n] \\ = & \frac{6 n - n^{2} - n}{2} \\ = & \frac{5 n - n^{2}}{2} \end{array}$

Evaluate the sum for n=100, n=500 and n=1000.

Substitute $100$ for $n$ in $\begin{array}{rcl} \frac{5 n - n^{2}}{2} \end{array}$ .

role="math" localid="1649054354093" $\begin{array}{rcl} \frac{5 (100) - {(100)}^{2}}{2} & = & \frac{500 - 10000}{2} \\ = & \frac{- 9500}{2} \\ = & - 4750 \end{array}$

Substitute $500$ for $n$ in $\begin{array}{rcl} \frac{5 n - n^{2}}{2} \end{array}$ .

$\begin{array}{rcl} \frac{5 (500) - {(500)}^{2}}{2} & = & \frac{2500 - 250000}{2} \\ = & - 123750 \end{array}$

Substitute $1000$ for $n$ in $\begin{array}{rcl} \frac{5 n - n^{2}}{2} \end{array}$ .

$\begin{array}{rcl} \frac{5 (1000) - {(1000)}^{2}}{2} & = & \frac{5000 - 1000000}{2} \\ = & - 497500 \end{array}$

Write the conclusion

The formula is $\begin{array}{rcl} \frac{5 n - n^{2}}{2} \end{array}$ .

The sum when $n = 100, 500$ and $1000$ is $- 4750, - 123750$ and $- 497500$ .

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Find a formula for each of the sums in Exercises 35, and then use these formulas to calculate each sum for $n = 100, n = 500$ and $n = 1000$ .
$\sum_{k = 1}^{n} (3 - k)$

Short Answer

Step by step solution

Given information

Determine the formula for the given summation.

Evaluate the sum for n=100, n=500 and n=1000.

Write the conclusion

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Find a formula for each of the sums in Exercises 35, and then use these formulas to calculate each sum for n=100,n=500and n=1000.∑k=1n(3-k)

Short Answer

Step by step solution

Given information

Determine the formula for the given summation.

Evaluate the sum for n=100, n=500 and n=1000.

Write the conclusion

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Discrete Mathematics

Mechanics Maths

Geometry

Statistics

Logic and Functions

Study anywhere. Anytime. Across all devices.

Find a formula for each of the sums in Exercises 35, and then use these formulas to calculate each sum for $n = 100, n = 500$ and $n = 1000$ .
$\sum_{k = 1}^{n} (3 - k)$