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Published byTerrance Harbinson Modified over 2 years ago

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Unit 1 Outcome 4 Recurrence Relations Sequences A591317……. B361224……. C235813…….. D ……… E235711……… In the above sequences some have obvious patterns while others don’t however this does not mean that a pattern doesn’t exist.

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Notation Suppose we write the terms of a sequence as u 1, u 2, u 3, …….., u n-1, u n, u n+1, ……... where u 1 is the 1 st term, u 2 is the 2 nd term etc…. and u n is the n th term ( n being any whole number.) The terms of a sequence can then be defined in two ways

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Either Using a formula for the nth term, u n, in terms of the value n. Or By expressing each term using the previous term(s) in the sequence. This is called a Recurrence Relation. Now reconsider the sequences at the start

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A591317……. Formula: u n = 4n + 1 So u 100 = 4 X = 401 Recurrence Relation: u n+1 = u n + 4 with u 1 = 5 So u 2 = u = = 9, u 3 = u = = 13, etc

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B361224……. Formula: u n = 3 X 2 n-1 So u 10 = 3 X 2 9 = 3 X 512 = 1536 Recurrence Relation: u n+1 = 2u n with u 1 = 3. So u 2 = 2u 1 = 2 X 3 = 6, u 3 = 2u 2 = 2 X 6 = 12, etc

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C235813…….. No formula this time but we have a special type of recurrence relation called a FIBONACCI SEQUENCE. Here u 1 = 2, u 2 = 3 then we have u 3 = u 2 + u 1 = = 5, u 4 = u 3 + u 2 = = 8, etc In general u n+2 = u n+1 + u n ie apart from 1st two, each term is the sum of the two previous terms.

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D ……… This sequence doesn’t have a recurrence relation but the terms can be found using the formula u n = n 3 - n + 17 Quite a tricky formula but it does work... u 1 = = 17 u 2 = = = 23 etc Also u 10 = = = 1007

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E235711……… This sequence is the PRIME NUMBERS (NB: Primes have exactly two factors !!) There is neither a formula nor a recurrence relation which will give us all the primes.

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