4.4 Sigma notation for sums — Practise

Work through these on your own. Check the start and the end of the index first. When the summand is a polynomial, the standard results are faster than adding the terms one by one.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Evaluate $\sum_{r=1}^{4}(5r - 2)$ .
Evaluate $\sum_{r=1}^{5}2^r$ .
Evaluate $\sum_{r=1}^{8}(3r - 2)$ .
Evaluate $\sum_{r=1}^{4}r^2$ .
Write $5 + 10 + 15 + \dots + 60$ in sigma notation, and work out the sum.
Use the standard results to work out $\sum_{r=1}^{12}(2r + 3)$ .
Use a standard result to work out $\sum_{r=1}^{10}r^2$ .
Use the standard results to work out $\sum_{r=1}^{8}(r^2 + r)$ .
Use a standard result to work out $\sum_{r=1}^{4}r^3$ .
Use the standard results to work out $\sum_{r=4}^{9}r$ .
The quiz night raises $50n + 100$ pounds in month $n$ . Write the total raised in the first 10 months in sigma notation, and work it out.
Work out $\sum_{r=1}^{n}(3r + 1)$ in terms of $n$ .

Answers and concise solutions

1. The sum

Answer: Write out the terms: $3 + 8 + 13 + 18 = 42$ . There are four terms, because the index runs from $1$ to $4$ . The terms rise by $5$ each time.

2. The sum

Answer: Write out the terms: $2 + 4 + 8 + 16 + 32 = 62$ . The summand is the term, not the sum, so each value of $r$ gives one power of $2$ . The terms double each time, which is why the total is close to the last term.

3. The sum

Answer: The terms are $1, 4, 7, 10, 13, 16, 19, 22$ . Adding them gives $92$ . The last term is $3(8) - 2 = 22$ , which confirms the upper limit. Pairing the ends works too: four pairs each totalling $23$ give $4 \times 23 = 92$ .

4. The sum

Answer: The terms are $1, 4, 9, 16$ , and their sum is $30$ . The standard result gives the same answer: $\frac{4 \times 5 \times 9}{6} = 30$ .

5. The sum

Answer: The terms are multiples of $5$ , so the rule is $5r$ . The last term is $60$ , and $5r = 60$ gives $r = 12$ , so the sum is $\sum_{r=1}^{12}5r = 5 \times \frac{12 \times 13}{2} = 5 \times 78 = 390$ . The upper limit $12$ is the position of the last term.

6. The sum

Answer: Split the sum: $\sum(2r + 3) = 2\sum r + \sum 3$ . Now $\sum_{r=1}^{12}r = \frac{12 \times 13}{2} = 78$ , and the constant $3$ is added twelve times, giving $36$ . So the sum is $2(78) + 36 = 156 + 36 = 192$ . The constant contributes $12 \times 3$ , not $3$ .

7. The sum

Answer: Use $\sum_{r=1}^{n}r^2 = \frac{n(n+1)(2n+1)}{6}$ with $n = 10$ : $\frac{10 \times 11 \times 21}{6} = \frac{2310}{6} = 385$ . Adding the terms by hand also gives $385$ , but the formula is quicker and less likely to slip.

8. The sum

Answer: Split the sum: $\sum(r^2 + r) = \sum r^2 + \sum r$ . With $n = 8$ , $\sum r^2 = \frac{8 \times 9 \times 17}{6} = 204$ and $\sum r = \frac{8 \times 9}{2} = 36$ . The total is $204 + 36 = 240$ . Splitting the sum is valid because each term is added separately.

9. The sum

Answer: Use $\sum_{r=1}^{n}r^3 = \left(\frac{n(n+1)}{2}\right)^2$ with $n = 4$ : $\left(\frac{4 \times 5}{2}\right)^2 = 10^2 = 100$ . The terms are $1, 8, 27, 64$ , and their sum is $100$ , so the result checks.

10. The sum

Answer: The sum starts at $4$ , not $1$ . Adjust by subtracting the missing part: $\sum_{r=4}^{9}r = \sum_{r=1}^{9}r - \sum_{r=1}^{3}r = \frac{9 \times 10}{2} - \frac{3 \times 4}{2} = 45 - 6 = 39$ . Check by adding: $4 + 5 + 6 + 7 + 8 + 9 = 39$ .

11. The total

Answer: The total is $\sum_{n=1}^{10}(50n + 100) = 50\sum_{n=1}^{10}n + \sum_{n=1}^{10}100 = 50 \times 55 + 1000 = 2750 + 1000 = 3750$ . The quiz night raises £3750 in the first 10 months. The constant $100$ is added once for each of the ten months.

12. The sum in terms of n

Answer: Split the sum: $3\sum_{r=1}^{n}r + \sum_{r=1}^{n}1 = 3 \times \frac{n(n+1)}{2} + n$ . Take out the common factor $n$ : $n\left(\frac{3(n+1)}{2} + 1\right) = \frac{n(3n + 5)}{2}$ . Check with $n = 5$ : $\frac{5 \times 20}{2} = 50$ , which matches the sum $4 + 7 + 10 + 13 + 16$ .