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Introduction • Chapter 5 of FP1 focuses on methods to calculate the sum of a series of numbers • The main focus is based around summing the first ‘n’ natural numbers of a given power • You will also become familiar with the proper series notation (you may already have seen this if you have covered Arithmetic and Geometric sequences from C1 and C2) Series You need to be able to understand and use the Σ notation This is the last value you put in to the formula (hence giving you the last term to be summed) This means ‘the sum of’ 10 5 + 1 =1 This is the first value you put in to the formula (hence giving you the first term to be summed) This is the last value you put in to the formula (hence giving you the last term to be summed) This is the formula for the sequence you are calculating the sum of (the ‘nth’ term) This means ‘the sum of’ 40 2 =0 This is the formula for the sequence you are calculating the sum of (the ‘nth’ term) This is the first value you put in to the formula (hence giving you the first term to be summed) 5A Series You need to be able to understand and use the Σ notation This is the last value you put in to the formula (hence giving you the last term to be summed) This means ‘the sum of’ 20 2 − 3 =10 This is the last value you put in to the formula (hence giving you the last term to be summed) This is the formula for the sequence you are calculating the sum of (the ‘nth’ term) This is the first value you put in to the formula (hence giving you the first term to be summed) In this example you are effectively calculating the sum of the terms from the 10th to the 20th This means ‘the sum of’ 4 − 5 =1 This is the formula for the sequence you are calculating the sum of (the ‘nth’ term) This is the first value you put in to the formula (hence giving you the first term to be summed) In this example you are not told how many terms there are, so the answer will be in terms of ‘n’, the number of terms… 5A Series You need to be able to understand and use the Σ notation Write out the terms defined by the following notation, and hence calculate the sum of the series: 10 (2 − 1) =1 Put r = 1 in for the first term, r = 2 in for the second and so on… Stop after calculating the term for r = 10 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 Write down the first 3 terms of, and the final term of, the sequence indicated by the notation below (2 + 1) =3 Put r = 3 in for the first term, r = 4 in for the second and so on… Put ‘n’ in for the final term… 7 + 9 + 11 + ……………………………………+ (2n + 1) = 100 5A Series You need to be able to understand and use the Σ notation Write the sequence below using the ∑ notation. 8 (3 − 1) =1 2 + 5 + 8 + 11 + 14 + 17 + 20 + 23 You need the formula for the sequence and the values to put in for the first and final terms! Formula for the sequence = ‘3n – 1’ (from GCSE maths! Substituting 1 in will get 2 – the first term Substituting 8 in will get 23 – the final term It is usually possible to do this several ways The notation below will give the same sequence (although is not as easy to work out!) 9 (3 − 4) =2 5A Series −2 You need to be able to understand and use the Σ notation ( − 1) Write the sequence below using the ∑ notation. =1 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + …… + (n – 1)(n – 2) You need the formula for the sequence and the values to put in for the first and final terms! Formula for the sequence = n(n+1) (multiply the term by the number one bigger than it! Substituting 1 in will get the first term To get the final term, we sub in (n – 2) ( + 1) ( − 2)( − 2 + 1) = ( − 2)( − 1) = ( − 1)( − 2) Replace n with ‘n – 2’ Simplify Rewrite This is the final term! 5A Series You need to be able to use the formula for the sum of the first n natural numbers The sum of the first ‘n’ natural numbers: 1 + 2 + 3 + 4 + …… + n is an arithmetic sequence with ‘n’ terms, with a = 1 and l (last term) = n =1 = +1 2 The formula for the sum of the first ‘n’ natural numbers is shown to the right: You DO NOT get given this formula on the exam! This is just saying the sum of the first ‘n’ natural numbers This is the formula for the sum of the first ‘n’ natural numbers The formula for the sequence is just ‘r’ as when you substitute in the term number, that is the term itself! For example if you wanted the sum of the first 30 numbers, let n = 30 and calculate the answer! 5B = Series You need to be able to use the formula for the sum of the first n natural numbers =1 = =1 Calculate the sum of the series indicated below: 50 = 50 =1 +1 2 50 50 + 1 2 We want the sum of the first 50 terms, so n = 50 Calculate 50 =1 +1 2 = 1275 =1 +1 2 ( + 1) 2 It is fine (and sometimes easier) to use the formula in this form! 5B Series =1 60 =21 = =1 +1 2 The sum of the numbers from 21 to 60… You need to be able to use the formula for the sum of the first n natural numbers Calculate the sum of the series indicated below: ( + 1) = 2 0 10 20 30 40 50 60 … Will be equal to the sum of the numbers from 1 to 60, subtract the numbers from 1 to 20… The notation will look like this… 60 This is asking you to find the sum of the numbers from 21 to 60 60 = =21 Sum from 21 to 60 20 − =1 Sum from 1 to 60 Notice the number here will always be one less than the one at the start! =1 Sum from 1 to 20 5B Series You need to be able to use the formula for the sum of the first n natural numbers Calculate the sum of the series indicated below: 60 60 = =21 = =1 +1 2 − =1 Sub in values for each part = =21 20 =1 60 60 =1 ( + 1) = 2 60 × 61 20 × 21 − 2 2 Calculate =21 = 1620 This is asking you to find the sum of the numbers from 21 to 60 So the sum of the numbers from 21 to 60 is 1620! 5B Series =1 ( + 1) = 2 = =1 +1 2 You need to be able to use the formula for the sum of the first n natural numbers Remember the link between the starting number and the sum we subtract! This is the general form for the problem you have seen (where you sum the numbers of a section of natural numbers, not starting on 1) () = = −1 () − =1 The sum of the numbers from 1 to n () =1 The sum of the numbers we will be removing (k – 1) 5B Series 2−1 = 2 2 − − 10 ≥3 = =1 − =5 =1 =1 Write out the formula twice as you will need it for both! = ( + 1) ( + 1) − 2 2 This type of question can look confusing but in reality you proceed as before = (2 − 1)(2 − 1 + 1) 4 × 5 − 2 2 We want the sum of the natural numbers from 5 to 2N – 1 = 20 (2 − 1)(2) − 2 2 =5 This will be the sum of the natural numbers from 1 to 2N – 1, subtract the numbers from 1 to 4. +1 2 4 2−1 = 2−1 Write out the formula for the numbers you want… You need to be able to use the formula for the sum of the first n natural numbers Show that: =1 ( + 1) = 2 4 2 − 2 − 20 = 2 Sub (2N – 1) in for the 1st, and 4 in for the 2nd Simplify or calculate where possible Expand brackets and group up Simplify = 2 2 − − 10 5B Series You need to be able to use the formula for the sum of the first n natural numbers Show that: 2−1 = 2 2 − − 10 ≥3 =5 =1 ( + 1) = 2 = =1 +1 2 But why N ≥ 3? Remember that you are told k = 5, meaning the first number in the sequence we are summing should be 5! If you use the value of N = 3, the sum of the sequence is 5, hence 3 is the lowest number we can put in! This type of question can look confusing but in reality you proceed as before We want the sum of the natural numbers from 5 to 2N – 1 This will be the sum of the natural numbers from 1 to 2N – 1, subtract the numbers from 1 to 4. 5B =1 ( + 1) = 2 = =1 +1 2 Series You need to be able to split up parts of a sequence and sum them separately You can split up series sums of the form: ( + ) =1 into 2 separate ‘series sums’ as follows: + =1 1 =1 This allows you to then use the sum formulae for the sequence overall! 5C =1 ( + 1) = 2 = =1 +1 2 Series You need to be able to split up parts of a sequence and sum them separately = 3 × 1 + 2 + 3× 2 + 2 + 3× 3 + 2 + 3 × 4 + 2 ……… (3 + 2) This is equal to the sum of the multiplied terms, added to the sum of the 2s =1 = 3 × 1 + 3 × 2 + 3 × 3 + 3 × 4 + 2+ 2 + 2 + 2 …… Can be written as: + 2 =1 =1 If we wrote out the first few terms of this sequence… Show that: 3 (3 + 2) 1 =1 We can ‘factorise’ the 3 out of the multiplied terms and factorise a 2 from the added terms… = 3 × 1 + 2 + 3 + 4 ……… + 2 1 + 1 + 1 + 1 ……… This is 3 multiplied by the sum of the first ‘n’ numbers represented by the formula ‘r’ This is 2 multiplied by ‘n’ 1s 3 + 2 =1 1 =1 5C =1 ( + 1) = 2 = =1 +1 2 Series 25 You need to be able to split up parts of a sequence and sum them separately (3 + 1) Split into two separate parts as you have seen =1 25 Evaluate: 3 25 + =1 (3 + 1) =1 = You need to split this up and sum the parts separately! = 25 1 =1 3( + 1) + (1) 2 3(25)(25 + 1) + 25(1) 2 Write the formulae for the sums. Remember the 3 at the start of the first one! We will also have ‘n’ lots of 1 Sub in n = 25 (25 terms to add up) Calculate = 1000 So the first 25 terms of the sequence with the formula (3r + 1) will add up to 1000! 5C =1 ( + 1) = 2 = =1 +1 2 Series You need to be able to split up parts of a sequence and sum them separately (7 − 4) =1 Show that: 7 − 1 (7 − 4) = 2 =1 7 = The sum of the first ‘n’ terms of this sequence Is given by this formula, where ‘n’ is the number of terms The two expressions are equivalent! − 4 =1 7( + 1) − 4 2 8 7( + 1) − 2 2 7 + 1 − 8 = 2 = In this case you should proceed as normal, but use ‘n’ instead! =1 = Split up as two separate sums 72 + 7 − 8 2 72 − = 2 (7 − 1) = 2 Remember the 7 on the first expression! We also have n lots of 4 Write ‘4n’ as fraction over 2 (for grouping) Group terms Expand the bracket Group terms Factorise 5C =1 ( + 1) = 2 = =1 +1 2 Series Write as one sum subtract another You need to be able to split up parts of a sequence and sum them separately Show that: (7 − 4) = =1 7 − 1 2 Hence, calculate the value of: 50 (7 − 4) 50 19 50 (7 − 4) = =20 (7 − 4) − =1 (7 − 4) =1 = (7 − 1) (7 − 1) − 2 2 = 50(7(50) − 1) 19(7(19) − 1) − 2 2 = 8725 − 1254 = 7471 Write the formula separately for each sum Sub 50 into the first and 19 into the second Calculate each Finish off! =20 Here, you can use the formula you’re given – remember that this will be the sum of the first 50 terms subtract the sum of the first 19! 5C Series You need to be able to calculate the sum of a sequence based on powers of 2 and 3 The sum of a sequence of squared numbers is given as follows: 2 = =1 ( + 1)(2 + 1) 6 ( + 1)(2 + 1) 6 And the formula for the sum of a sequence of cubes is: 2 + 1 = 4 =1 3 2 2 +1 4 You will see where these come from in chapter 6! 2 You get given these formulae in these forms in the exam booklet! (Remember you do not get the formula for a linear sequence!) 5D =1 ( + 1) = 2 =1 2 + 1 4 2 30 2 =1 Write out the formula for a squared sequence = ( + 1)(2 + 1) 6 = (30)(30 + 1)(2(30) + 1) 6 = (30)(31)(61) 6 30 2 3 = Series =1 You need to be able to calculate the sum of a sequence based on powers of 2 and 3 Evaluate: ( + 1)(2 + 1) = 6 2 =1 Sub in n = 30 as we want 30 terms Simplify the numerator (if necessary!) Calculate = 9455 5D =1 ( + 1) = 2 You need to be able to calculate the sum of a sequence based on powers of 2 and 3 ( + 1)(2 + 1) = 6 2 3 = Series =1 =1 40 40 3 = 19 3 − =1 3 =1 Evaluate: 3 2 + 1 = 4 =20 Remember for this one you need the sum of the first 40 terms, subtract the first 19 terms! 2 Write it as one sum subtract another =20 40 2 + 1 4 = 2 2 + 1 − 4 (40)2 40 + 1 4 2 − 2 (19)2 19 + 1 4 Write out the formula for the cubed sequence twice 2 Sub in 40 for the first and 19 for the second Calculate = 672400 − 36100 Finish the sum! = 636300 5D =1 ( + 1) = 2 2 You need to be able to calculate the 2 sum of a sequence based on powers of =+1 2 and 3 =+1 2 + 1 4 2 2 − = =1 2 Write out the formula twice =1 ( + 1)(2 + 1) ( + 1)(2 + 1) − 6 6 = 2(2 + 1)(4 + 1) ( + 1)(2 + 1) − 6 6 This one is more algebraic but you still approach it the same way! = The first value we put in the sequence will be ‘n + 1’ = (2 + 1) 2 4 + 1 − ( + 1) 6 The final value we put in will be ‘2n’ = (2 + 1) 8 + 2 − − 1 6 So we want the sum of the first ‘2n’ terms, subtract the first ‘n’ terms (same as if we were using numbers!) =1 = 2 2 3 = Series =1 2 Find: ( + 1)(2 + 1) = 6 2 2 2 + 1 4 + 1 − ( + 1)(2 + 1) 6 [ [ = (2 + 1)(7 + 1) 6 ] ] Sub ‘2n’ into the first and ‘n’ into the second You can write this as one fraction This is the key step – you can factorise as n(2n+1) is common to both terms! Expand the terms in the square bracket Simplify the square bracket (which can now be written as a ‘normal’ bracket!) The factorising step is crucial here – otherwise you will end up trying to factorise a cubic which can take a long time! 5D =1 ( + 1) = 2 ( + 1)(2 + 1) = 6 2 Series =1 2 You need to be able to calculate the sum of a sequence based on powers of 2 and 3 2 = =+1 3 = =1 2 + 1 4 2 (2 + 1)(7 + 1) 6 If n = 1 Find: 2 2 2 2 =+1 = (2 + 1)(7 + 1) 6 Verify that the result is correct for n = 1, 2 and 3 (This can show the formula is working, although in reality isn’t a proof in itself!) =2 The first number we put in is 2, which is also the last number we put in Sequence 4 (2 + 1)(7 + 1) 6 = 1(2 + 1)(7 + 1) 6 = 24 6 =4 So the numbers in the sequence just add up to 4! Let’s check the formula! 5D =1 ( + 1) = 2 ( + 1)(2 + 1) = 6 2 Series =1 2 You need to be able to calculate the sum of a sequence based on powers of 2 and 3 2 = =+1 3 = =1 2 + 1 4 2 (2 + 1)(7 + 1) 6 If n = 2 Find: 2 4 2 2 =+1 = (2 + 1)(7 + 1) 6 Verify that the result is correct for n = 1, 2 and 3 (This can show the formula is working, although in reality isn’t a proof in itself!) =3 The first number we put in is 3, and the last number we put in in 4 Sequence 9, 16 (2 + 1)(7 + 1) 6 = 2(4 + 1)(14 + 1) 6 = 150 6 = 25 So the numbers in the sequence add up to 25 Let’s check the formula! 5D =1 ( + 1) = 2 ( + 1)(2 + 1) = 6 2 3 = Series =1 2 You need to be able to calculate the sum of a sequence based on powers of 2 and 3 2 = =+1 =1 2 + 1 4 2 (2 + 1)(7 + 1) 6 If n = 3 Find: 2 6 2 2 =+1 = (2 + 1)(7 + 1) 6 Verify that the result is correct for n = 1, 2 and 3 (This can show the formula is working, although in reality isn’t a proof in itself!) =4 The first number we put in is 4, and the last number we put in in 6 Sequence 16, 25, 36 So the numbers in the sequence add up to 77 Let’s check the formula! (2 + 1)(7 + 1) 6 = 3(6 + 1)(21 + 1) 6 = 462 6 = 77 So the formula seems to be working fine! 5D =1 ( + 1) = 2 ( + 1)(2 + 1) = 6 2 =1 3 = 2 + 1 4 2 Series =1 You need to be able to use all you have learnt to calculate the sum of a more complex series, made up of several terms As you saw in section 5C, you can take out a coefficient of a term in order to sum it. You can also do this with the sums for r2 and r3. For example: 4 2 = 4 =1 2 =1 = 4( + 1)(2 + 1) 6 You need to remember to include the coefficient in the formula though! 5E =1 ( + 1) = 2 ( + 1)(2 + 1) = 6 2 =1 3 = 2 + 1 4 2 Series =1 You need to be able to use all you have learnt to calculate the sum of a more complex series, made up of several terms 2 + − 2 =1 2 Show that: 2 + − 2 = =1 = = = ( + 4)( − 1) 3 2(2 + 3 − 4) 6 (2 + 3 − 4) 3 ( + 4)( − 1) 3 Divide numerator and denominator by 2 + − =1 Write as separate sums =1 2 =1 = ( + 1) ( + 1)(2 + 1) − 2 + 2 6 = 12 ( + 1)(2 + 1) 3( + 1) − + 6 6 6 + 1 2 + 1 + 3 + 1 − 12 = 6 = Factorise! = = = [ ( + 1)(2 + 1) + 3( + 1) − 12 ] 6 [ 2 2 + 3 + 1 + 3 + 3 − 12 6 (22 + 6 − 8) 6 2 (2 + 3 − 4) 6 ] Write the formula for each part in terms of n Write all with a common denominator Group up ‘Clever factorisation’ Expand brackets Group terms Take the factor 2 out of the bracket 5E =1 ( + 1) = 2 ( + 1)(2 + 1) = 6 2 =1 3 = 2 + 1 4 2 Series =1 You need to be able to use all you have learnt to calculate the sum of a more complex series, made up of several terms 2 + − 2 =1 =1 1 2 + 1 −2 =0 =2 2 2 + 2 −2 =4 =3 3 2 + 3 −2 = 10 4 + 10 + 18 + 28 + 40 … … … + 418 =4 4 2 + 4 −2 = 18 You can see that this formula gives us the sequence we are trying to find the sum of! =? Show that: 2 + − 2 = =1 ( + 4)( − 1) 3 Hence, calculate the sum of the series: (The 0 at the start will not affect the sum so can be ignored!) 2 + − 2 = 418 We need to know how many terms there are, so have to find the value for r which gives a term with a value of 418… 5E =1 ( + 1) = 2 ( + 1)(2 + 1) = 6 2 =1 3 = 2 Series =1 You need to be able to use all you have learnt to calculate the sum of a more complex series, made up of several terms Show that: 2 + 1 4 2 + − 2 = 418 + − 420 = 0 ( − 20)( + 21) = 0 + − 2 = ( + 4)( − 1) 3 2 =1 Hence, calculate the sum of the series: 4 + 10 + 18 + 28 + 40 … … … + 418 Subtract 418 2 Factorise 2 answers, only 1 is possible though! = 20 − 21 So we are finding the sum of the first 20 terms of the sequence! 20 2 + − 2 =1 = ( + 4)( − 1) 3 20 = (20 + 4)(20 − 1) 3 = 3040 We can use the formula we were given! Sub in n = 20 Calculate 5E =1 ( + 1) = 2 ( + 1)(2 + 1) = 6 2 =1 3 = 2 + 1 4 2 Series =1 2 You need to be able to use all you have learnt = 2 + 1 4 to calculate the sum of a more complex series, made up of several terms 2 32 + 1 = 6 2 Find a formula for the sum of the series: + 3( + 1) 5( + 1)(2 + 1) − 2 6 + 5( + 1)(2 + 1) 9( + 1) − 6 6 Combine ( + 3)(2 − 1) =1 Expand the bracket (2 + 5 − 3) =1 Expand the bracket again 2 3 + 5 2 − 3 =1 2 + 5 =1 = 22 + 1 4 =1 2 + Write as 3 separate sums 3 2 = = 2 − 3 =1 3( + 1) 5( + 1)(2 + 1) − 2 6 Write with the same denominator = = 32 + 1 2 + 5 + 1 2 + 1 − 9( + 1) 6 [ ( + 1) 3( + 1) + 5(2 + 1)− 9 ] 6 [ 2 ( + 1) 3 + 3 + 10 + 5 − 9 6 ( + 1)(32 + 13 − 4) 6 Write using the formulae above. Remember to include the coefficients! ] ‘Clever factorisation’ Expand the inner brackets Simplify (you should also factorise if possible) 5E =1 ( + 1) = 2 ( + 1)(2 + 1) = 6 2 =1 3 = Find a formula for the sum of the series: Write as one sum subtract another 40 40 ( + 3)(2 − 1) = =11 ( + 3)(2 − 1) =1 =11 ( + 3)(2 − 1) − =1 ( + 3)(2 − 1) =1 = ( + 1)(32 + 13 − 4) ( + 1)(32 + 13 − 4) − 6 6 = 10(11)(426) 40(41)(5316) − 6 6 40 ( + 3)(2 − 1) 10 Write the formulae out twice, one for each sum! Hence, calculate the following: 2 Series =1 You need to be able to use all you have learnt to calculate the sum of a more complex series, made up of several terms ( + 1)(32 + 13 − 4) = 6 2 + 1 4 Sub in 40 and 10 Calculate! = 1445230 5E Summary • We have seen how to calculate the sum of a series in various circumstances • We have practiced the correct series notation • We have also seen and used the ‘clever factorisation’ method for simplifying expressions!