FP2 Chapter 2 – Method of Differences Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 3rd August 2015

STARTER – Round 1! 1 2 × 2 3 × 3 4 ×…× 𝑛−1 𝑛 = 𝟏 𝒏 ? ? 1− 1 4 1− 1 9 1− 1 16 … 1− 1 𝑛 2 = 𝟏− 𝟏 𝟐 𝟏+ 𝟏 𝟐 𝟏− 𝟏 𝟑 𝟏+ 𝟏 𝟑 … 𝟏− 𝟏 𝒏 𝟏+ 𝟏 𝒏 = 𝟏 𝟐 × 𝟑 𝟐 × 𝟐 𝟑 × 𝟒 𝟑 × 𝟑 𝟒 × 𝟓 𝟒 ×…× 𝒏−𝟏 𝒏 × 𝒏+𝟏 𝒏 = 𝒏+𝟏 𝟐𝒏 ? ?

STARTER – Round 2! Bro Hint: Perhaps write out the first few terms? 𝑟=1 𝑛 1 𝑟 − 1 𝑟+1 = 𝟏 𝟏 − 𝟏 𝟐 + 𝟏 𝟐 − 𝟏 𝟑 +…+ 𝟏 𝒏 − 𝟏 𝒏+𝟏 =𝟏− 𝟏 𝒏+𝟏 = 𝒏 𝒏+𝟏 ? ! If 𝑢 𝑛 =𝑓 𝑛 −𝑓 𝑛+1 then 𝑟=1 𝑛 𝑢 𝑟 =𝑓 1 −𝑓 𝑛+1 Known as ‘method of differences’.

Example ? ? ? ? Show that 4 𝑟 3 = 𝑟 2 𝑟+1 2 − 𝑟−1 2 𝑟 2 Hence prove, by the method of differences that 𝑟=1 𝑛 𝑟 3 = 1 4 𝑛 2 𝑛+1 2 Bro Exam Note: Exam questions usually have two parts: Showing some expression is equivalent to one in form 𝑓 𝑛 −𝑓(𝑛+1) Using method of differences to simplify summation. 𝑟 2 𝑟+1 2 − 𝑟−1 2 𝑟 2 = 𝑟 2 𝑟 2 +2𝑟+1 − 𝑟 2 𝑟 2 −2𝑟+1 =…=4 𝑟 3 =𝐿𝐻𝑆 𝑟=1 𝑛 𝑟 3 = 1 2 2 2 − 0 2 1 2 + 2 2 3 2 − 1 2 2 2 + 3 2 4 2 − 2 2 3 2 +…+ 𝑛 2 𝑛+1 2 − 𝑛−1 2 𝑛 2 = 𝑛 2 𝑛+1 2 Therefore 4 𝑟=1 𝑛 𝑟 3 = 𝑛 2 𝑛+1 2 So 𝑟=1 𝑛 𝑟 3 = 𝑛 2 𝑛+1 2 ? ? Carefully look at the pattern of cancelling. The second term in each pair is cancelled out by the first term in the previous pair. Thus in the last pair the first term won’t have been cancelled out. While not essential, I like to put brackets around pairs to see the exact matchings more easily. ? ?

Partial Fraction Example Find 𝑟=1 𝑛 1 4 𝑟 2 −1 using the method of differences. First split into partial fractions in hope that we get form 𝑓 𝑟 −𝑓 𝑟+1 or something similar. 1 2𝑟+1 2𝑟−1 = 𝐴 2𝑟+1 + 𝐵 2𝑟−1 𝐴=− 1 2 , 𝐵= 1 2 Thus 𝑟=1 𝑛 1 4 𝑟 2 −1 = 1 2 𝑟=1 𝑛 1 2𝑟−1 − 1 2𝑟+1 = 1 2 1 1 − 1 3 + 1 3 − 1 5 + 1 5 − 1 7 +…+ 1 2𝑛−1 − 1 2𝑛+1 = 1 2 1− 1 2𝑛+1 = 1 2 2𝑛+1−1 2𝑛+1 = 𝑛 2𝑛+1 ? ? ? ? ? ?

Test Your Understanding FP2 June 2013 Q1 (a) Express 2 2𝑟+1 2𝑟+3 in partial fractions. (b) Using your answer to (a), find, in terms of 𝑛, 𝑟=1 𝑛 3 2𝑟+1 2𝑟+3 Give your answer as a single fraction in its simplest form. ?

Harder Ones: 𝑓 𝑟 −𝑓 𝑟+2 Usually in exams, they try to make it slightly harder by using the form 𝑓 𝑟 −𝑓(𝑟+2) instead of 𝑓 𝑟 −𝑓 𝑟+1 . The result is that terms don’t cancel in adjacent pairs, but in pairs further away. You just have to be really really careful (really) when you see what terms cancel. 𝑟=1 𝑛 1 𝑟 − 1 𝑟+2 = 1 1 − 1 3 + 1 2 − 1 4 + 1 3 − 1 5 + 1 4 − 1 6 …+ 1 𝑛 − 1 𝑛+2 = 1 1 + 1 2 − 1 𝑛+1 − 1 𝑛+2 = 𝑛 3𝑛+5 2 𝑛+1 𝑛+2 ? ? We can see the second term of each pair is cancelled out by the first term of two pairs later. This means the first term of the first two pairs won’t be cancelled, and the second term of the last two pairs won’t cancel. ?

Test Your Understanding FP2 June 2013 (R) Q3 Express 2 𝑟+1 𝑟+3 in partial fractions. Hence show that: 𝑟=1 𝑛 2 𝑟+1 𝑟+3 = 𝑛 5𝑛+13 6 𝑛+2 𝑛+3 Evaluate 𝑟=10 100 2 𝑟+1 𝑟+3 , giving your answer to 3 significant figures. ? ? ?

Exercise 2 ? ? 1 2 6 Then move on to provided exam questions Show that 𝑟= 1 2 (𝑟 𝑟+1 −𝑟 𝑟−1 Hence show 𝑟=1 𝑛 𝑟 = 𝑛 2 𝑛+1 using the method of differences. Given 1 𝑟 𝑟+1 𝑟+2 ≡ 1 2𝑟 𝑟+1 − 1 2 𝑟+1 𝑟+2 find 𝑟=1 𝑛 1 𝑟 𝑟+1 𝑟+2 using the method of differences. 𝒏 𝒏+𝟑 𝟒 𝒏+𝟏 𝒏+𝟐 Given that 𝑟 𝑟+1 ! ≡ 1 𝑟! − 1 𝑟+1 ! find 𝑟=1 𝑛 𝑟 𝑟+1 ! 𝟏− 𝟏 𝒏+𝟏 ! 1 2 ? 6 ? Then move on to provided exam questions (the textbook here is pretty inadequate preparation for the exam, as it barely practices cancelling between non-adjacent pairs – i.e. most exam questions!)