FM Series

𝑥 2 +5𝑥+6 2𝑥(𝑥+3)+7(𝑥+3) 4 (𝑥+3) 2 −5(𝑥+3) Series KUS objectives BAT Understand and write series using sigma notation BAT Generate sequences BAT Solve arithmetic series problems involving sigma notation Starter factorise: 𝑥 2 +5𝑥+6 2𝑥(𝑥+3)+7(𝑥+3) 4 (𝑥+3) 2 −5(𝑥+3)

1 𝑛 (2𝑟−1) The sum of odd numbers is Notes on notation The sum of odd numbers is Sn = 1 + 3 + 5 + 7 + 9 + …. + (2n – 1) We could write this as: number of terms, the last term will be 2(n)-1 1 𝑛 (2𝑟−1) r is the position of each term in the series 1 is the first position number, so this series starts at 2(1) -1 = 1  is sigma in the ancient Greek alphabet and stands for ‘the sum of’

(Series are not always arithmetic) WB1 Try these: (Series are not always arithmetic) 1 𝑛 (3𝑟+5) = 8 + 11 + 14 + 17 + 20 + …. + (3n+5) 1 10 (3𝑟+5) = 8 + 11 + 14 + 17 + 20 + …. + 35 8 20 (3𝑟+5) = 29 + 32 + 35 + 38 + 41 + …. + 65 1 6 (3 𝑟 2 +1) = 4 + 13 + 28 + 49 + 76 + 109 1 20 (4𝑟−1) 3 + 7 + 11 + 15 + 19 + …. + 79 = 1 100 𝑟 2 1 + 4 + 9 + 16 + 25 + …. + 10000 = 1 25 (83−7𝑟) 76 + 69 + 62 + 55 + 48 + …. - 92 = 1 7 2 𝑟 3 2 + 16 + 54 + 128 + 250 + 432 + 686 = All these series start at the first term

For an arithmetic series the ‘sum of terms’ is worked out Notes on Sn For an arithmetic series the ‘sum of terms’ is worked out using this formula 𝑆𝑛 = 𝑛 2 2𝑎+ 𝑛−1 𝑑 Where: a is the first term, d is the common difference and n is the number of terms. We assume the sequence starts at the first term If an arithmetic series is the sum of the same constant term each time Then 𝑆𝑛 =𝑛𝑎 or 1 𝑛 𝑎 =𝑛𝑎

WB2 a) Evaluate 1 22 (3𝑟+5) using the formula for the sum of an arithmetic series 1 22 (3𝑟+5) = 8 + 11 + 14 + 17 + 20 + …. + (3n+5) 8 + 11 + 14 + 17 + 20 + …. + 71 This is an arithmetic series, using the formula for sum of terms 𝑆 𝑛 = n 2 2𝑎+ 𝑛−1 𝑑 a = 8 d = 3 n = 22 1 22 (3𝑟+5) = 22 2 16+21×3 = 869

This is an arithmetic series, using the formula for sum of terms WB2 b) Evaluate 1 40 (7−2𝑟) using the formula for the sum of an arithmetic series. Explain why the answer is negative 1 40 (7−2𝑟) = 5 + 3 + 1 + (-1) + (-3) + …. + (7-2(40)) This is an arithmetic series, using the formula for sum of terms 𝑆 𝑛 = n 2 2𝑎+ 𝑛−1 𝑑 1 40 (7−2𝑟) = 40 2 10+39×(−2) a = 5 d = -2 n = 40 =−1360 This negative because most of the terms added up are negative

This is NOT an arithmetic series, WB2 c) Evaluate 1 6 𝑟(𝑟−6) using the formula for the sum of an arithmetic series This is NOT an arithmetic series, You cannot use the formula for sum of terms 1 6 𝑟(𝑟−6) = 1(1-6) + 2(2-6) + 3(3-6) +4(4-6) + 5(5-6) + 0 = (-5) + (-8) + (-9) + (-8) + (-5)+0 = - 35

1 𝑛 (3𝑟+4) =3 1 𝑛 𝑟 +4𝑛 Show that: 7 + 10 + 13 + 16 + 19 + …. + (3n+4) WB3 key result 1 𝑛 (3𝑟+4) =3 1 𝑛 𝑟 +4𝑛 Show that: 1 𝑛 (3𝑟+4) = 7 + 10 + 13 + 16 + 19 + …. + (3n+4) Split into sums = (3x1+ 4) + (3x2+ 4) + (3x3+ 4) + (3x4+ 4)+ … + (3n+ 4) Split further = (3x1+ 3x2 + 3x3 + 3x4 +… + 3n) + (4 + 4+ 4+ 4+ … + 4) Split further still = 3 (1+ 2 + 3 + 4 +… + n) + (4 + 4+ 4+ 4+ … + 4) = 3 1 𝑛 𝑟 +4𝑛 𝑄𝐸𝐷

We will come back to this later WB3 (cont) 1 𝑛 4 =4𝑛 Note that: 1 𝑛 (𝑎𝑟+𝑏) =𝑎 1 𝑛 𝑟 +𝑏𝑛 and: We will come back to this later

Notes three key results Use Sn = ½n[2a+(n-1)d] to find a formula in terms of n for the sum of the first n natural numbers 1 𝑛 𝑟 = 𝑛 2 2×1+(𝑛−1)×1 = 𝑛 2 𝑛+1 The formula for the sum of the first n squares is 1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1) The formula for the sum of the first n cubes is 1 𝑛 𝑟 3 = 𝑛 2 4 (𝑛+1) 2 These are in the exam booklet and proofs are not required

Try these: ½ (50)(51) = 1275 2718 1/6 (20)(21)(41) = ¼ (14)2 (15)2 = 1 𝑛 𝑟 = 𝑛 2 𝑛+1 1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1) 1 𝑛 𝑟 3 = 𝑛 2 4 (𝑛+1) 2 WB4 Try these: 1 50 𝑟 = ½ (50)(51) = 1275 1 20 𝑟 2 = 2718 1/6 (20)(21)(41) = 1 14 𝑟 3 = ¼ (14)2 (15)2 = 11025

Remember: and: Calculate: =7[ 46 2 (46+1)]−3(46) =7429 WB5a 1 𝑛 4 =4𝑛 1 𝑛 𝑟 = 𝑛 2 𝑛+1 1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1) 1 𝑛 𝑟 3 = 𝑛 2 4 (𝑛+1) 2 WB5a 1 𝑛 4 =4𝑛 Remember: 1 𝑛 (𝑎𝑟+𝑏) =𝑎 1 𝑛 𝑟 +𝑏𝑛 and: Calculate: 1 46 (7𝑟−3) = 7 1 46 𝑟 −3(46) =7[ 46 2 (46+1)]−3(46) =7429

=3 (25)(26) 2 +25 =1000 Calculate 1 25 𝑟= 3 1 25 𝑟 + 25(1) 1 25 (3𝑟+1) 1 𝑛 𝑟 = 𝑛 2 𝑛+1 1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1) 1 𝑛 𝑟 3 = 𝑛 2 4 (𝑛+1) 2 WB5b 1 25 (3𝑟+1) Calculate 1 25 𝑟= 3 1 25 𝑟 + 25(1) =3 (25)(26) 2 +25 =1000

1 𝑛 𝑟 = 𝑛 2 𝑛+1 1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1) 1 𝑛 𝑟 3 = 𝑛 2 4 (𝑛+1) 2 WB 5c Calculate 1 28 30−5𝑟 1 28 30−5𝑟 =30×28−5 1 28 𝑟 =30×28−5 28 2 29 =−1190

WB6ab Find a) 50 100 𝑟 b) 10 25 𝑟 3 c) 5 30 ( 𝑟 2 −2) d) 11 25 𝑟(𝑟+1) 1 𝑛 𝑟 = 𝑛 2 𝑛+1 1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1) 1 𝑛 𝑟 3 = 𝑛 2 4 (𝑛+1) 2 WB6ab Find a) 50 100 𝑟 b) 10 25 𝑟 3 c) 5 30 ( 𝑟 2 −2) d) 11 25 𝑟(𝑟+1) A neat trick is: 𝑎) 50 100 𝑟 = 1 100 𝑟 − 1 49 𝑟 = 100 2 101 − 49 2 50 = 3825 𝑏) 10 25 𝑟 3 = 1 25 𝑟 3 − 1 9 𝑟 3 = 25 2 4 (26) 2 − 9 2 4 10 2 =103600

WB6c Find a) 50 100 𝑟 b) 10 25 𝑟 3 c) 5 30 ( 𝑟 2 −2) d) 11 25 𝑟(𝑟+1) 1 𝑛 𝑟 = 𝑛 2 𝑛+1 1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1) 1 𝑛 𝑟 3 = 𝑛 2 4 (𝑛+1) 2 WB6c Find a) 50 100 𝑟 b) 10 25 𝑟 3 c) 5 30 ( 𝑟 2 −2) d) 11 25 𝑟(𝑟+1) 𝑐) 5 30 𝑟 2 −2 = 1 30 𝑟 2 −2 − 1 4 𝑟 2 −2 = 30 6 31 61 −2×30 − 4 6 5 9 −2×4 = 9395 - 22 = 9373

WB6d Find a) 50 100 𝑟 b) 10 25 𝑟 3 c) 5 30 ( 𝑟 2 −2) d) 11 25 𝑟(𝑟+1) 1 𝑛 𝑟 = 𝑛 2 𝑛+1 1 𝑛 𝑟 2 = 𝑛 6 𝑛+1 (2𝑛+1) 1 𝑛 𝑟 3 = 𝑛 2 4 (𝑛+1) 2 WB6d Find a) 50 100 𝑟 b) 10 25 𝑟 3 c) 5 30 ( 𝑟 2 −2) d) 11 25 𝑟(𝑟+1) 𝑑) 11 25 𝑟(𝑟+1) = 1 25 𝑟 2 +𝑟 − 1 10 𝑟 2 +𝑟 = 25 6 26 51 + 25 2 26 − 10 6 11 21 + 10 2 11 = 5850 − 440 = 5410

WB7ab Try these: 1 10 (𝑟+1)(𝑟−2) =245: 1 20 𝑟 2 (𝑟−1) = 41230

Crucial points 1. Check your results When you find a sum of the first n terms of a series, it is a good idea to substitute n = 1, and perhaps n = 2 as well, to check your result. 2. Look for common factors When using standard results, there can be quite a lot of algebra involved in simplifying the result. Make sure you take out any common factors first, as this makes the algebra a lot simpler. 3. Be careful when summing a constant term remember 1 𝑛 𝑘 =𝑘+𝑘+𝑘+......+𝑘=𝑘𝑛

KUS objectives BAT Understand and write series using sigma notation BAT Generate sequences BAT Solve arithmetic series problems involving sigma notation self-assess One thing learned is – One thing to improve is –

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