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Arithmetic Series 19 May 2011. Summations Summation – the sum of the terms in a sequence {2, 4, 6, 8} → 2 + 4 + 6 + 8 = 20 Represented by a capital Sigma.

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Presentation on theme: "Arithmetic Series 19 May 2011. Summations Summation – the sum of the terms in a sequence {2, 4, 6, 8} → 2 + 4 + 6 + 8 = 20 Represented by a capital Sigma."— Presentation transcript:

1 Arithmetic Series 19 May 2011

2 Summations Summation – the sum of the terms in a sequence {2, 4, 6, 8} → 2 + 4 + 6 + 8 = 20 Represented by a capital Sigma

3 Summation Notation Sigma (Summation Symbol) Upper Bound (Ending Term #) Lower Bound (Starting Term #) Sequence

4 Example #1

5 Example #2

6 Example #3

7 Your Turn: Find the sum:

8

9

10 Partial Sums of Arithmetic Sequences – Formula #1 Good to use when you know the 1 st term AND the last term # of terms 1 st term last term

11 Formula #1 – Example #1 Find the partial sum: k = 9, u 1 = 6, u 9 = –24

12 Formula #1 – Example #2 Find the partial sum: k = 6, u 1 = – 4, u 6 = 14

13 Formula #1 – Example #3 Find the partial sum: k = 10, u 1 = 0, u 10 = 30

14 Your Turn: Find the partial sum: 1. k = 8, u 1 = 7, u 8 = 42 2. k = 5, u 1 = –21, u 5 = 11 3. k = 6, u 1 = 16, u 6 = –19

15 Partial Sums of Arithmetic Sequences – Formula #2 Good to use when you know the 1 st term, the # of terms AND the common difference # of terms 1 st termcommon difference

16 Formula #2 – Example #1 Find the partial sum: k = 12, u 1 = –8, d = 5

17 Formula #2 – Example #2 Find the partial sum: k = 6, u 1 = 2, d = 5

18 Formula #2 – Example #3 Find the partial sum: k = 7, u 1 = ¾, d = –½

19 Your Turn: Find the partial sum: 1. k = 4, u 1 = 39, d = 10 2. k = 5, u 1 = 22, d = 6 3. k = 7, u 1 = 6, d = 5

20 Choosing the Right Partial Sum Formula Do you have the last term or the constant difference?

21 Examples Identify the correct partial sum formula: 1. k = 6, u 1 = 10, d = –3 2. k = 12, u 1 = 4, u 12 = 100

22 Your Turn: Identify the correct partial sum formula and solve for the partial sum 1. k = 11, u 1 = 10, d = 2 2. k = 10, u 1 = 4, u 10 = 22 3. k = 16, u 1 = 20, d = 7 4. k = 15, u 1 = 20, d = 10 5. k = 13, u 1 = –18, u 13 = –102


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