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FP1 Chapter 5 - Series

FP1 Chapter 5 - Series. Dr J Frost (jfrost@tiffin.kingston.sch.uk) . Last modified: 8 th February 2013. Recap. A series is just a sequence, which can be either finite or infinite. Euler introduced the  symbol (capital sigma) to mean the sum of a series. What we’ll cover:

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FP1 Chapter 5 - Series

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  1. FP1 Chapter 5 - Series Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 8th February 2013

  2. Recap A series is just a sequence, which can be either finite or infinite. Euler introduced the  symbol (capital sigma) to mean the sum of a series. What we’ll cover: The sum of the first integers. The sum of the first squares. The sum of the first cubes. Breaking down more complex sums. Dealing with other bounds. We’ll later prove some of these in Chapter 6.

  3. Recap Determine the following results by explicitly writing out the elements in the sum. ? ? ?

  4. Sum of ones, integers, squares, cubes These are the four essential formulae you need to learn for this chapter (and that’s about it!): The last two are in the formula booklet, but you should memorise them anyway) ! ? Sum of first n integers ? Sum of first n squares ? Note that: Sum of first n cubes ? i.e. The sum of the first n cubes is the same as the square of the first n integers.

  5. QuickfireTriangulars! In your head... 1 + 2 + 3 + ... + 10 = 55 1 + 2 + 3 + ... + 99 = 4950 11 + 12 + 13 + ... + 20 = 210 – 55 = 155 100 + 101 + 102 + ... + 200 = 20100 – 4950 = 15150 ? ? ? ?

  6. Practice Use the formulae to evaluate the following. ? ? ? ? Bro Tip: Ensure that you use one less than the lower limit. ?

  7. Test Your Understanding Show that ?

  8. Breaking Up Summations ? Examples: ? ? ?

  9. Breaking Up Summations We can combine this property of summations with the previous one to break summations up. ? ? Examples: ? ?

  10. Past Paper Question ? Edexcel June 2013 ?

  11. Exercises Exercise 5E All questions Bonus Frostie Conundrum Given that n is even, determine 12 – 22 + 32 – 42 + 52 + ... – n2. ? Alternatively, notice we have pairs of difference of two pairs. We thus get: (3 × -1) + (7 × -1) + (11 × -1) + ... + ([2n-1] × – 1) = -1(3 + 7 + 11 + [2n – 1]) The contents of the brackets are the sum of an arithmetic series (with a = 3, d = 4, and n/2 terms), and we could get the same result.

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