Proof by Induction

Starter: write expressions for FM proof by induction: KUS objectives BAT understand and use the principle of proof by induction BAT prove results about sums of series Starter: write expressions for 1 𝑛 𝑟 = 𝑛 2 (𝑛+1) 1 𝑛+1 𝑟 2 = 1 6 (𝑛+1)(𝑛+2)(2𝑛+3) 11 𝑛+1 𝑟 = 1 2 𝑛+1 𝑛+2 −30

The sum of n positive odd numbers is n2 WB A1 General statement The sum of n positive odd numbers is n2 n = 1 n = 2 n = 3 n = 4 n = 5 We can see this is true but how can we prove it? Discuss your ideas

The way ‘proof by mathematical induction’ works is often likened to knocking dominoes over If the dominoes are lined up, then you knock over the first one, every domino afterwards will fall down iPhone Dominoes

Notes Proof by mathematical Induction 1. Basis Prove the statement is true for n = 1 2. Assumption That the statement is true for n = k 3. Inductive Show that the statement is then true for n = k + 1 4. Conclusion That the statement is then true for all positive integers, n Note that proof by induction is used to check if a given statement is true This is not the same as proving from first principles

The sum of n positive odd numbers is n2 WB A1 The sum of n positive odd numbers is n2 n = 1 True for n = 1 n = k Assume n = k +1 Discuss – the idea is to know what you are working towards Encourage students to use the details – true for n = 1, assume true, true for n = k+ 1 et cetera True for n = k+1 and for n=1 So True for all integers

Prove the result for the sum of n positive cube numbers: WB A2 General statement 1 𝑛 𝑟 3 = 1 4 𝑛 2 𝑛+1 2 Prove the result for the sum of n positive cube numbers: n = 1 True for n = 1 n = k Assume n = k +1 Discuss – the idea is to know what you are working towards Encourage students to use the details – true for n = 1, assume true, true for n = k+ 1 et cetera True for n = k+1 So True for all integers

1 𝑛 𝑟 2 = 1 6 𝑛 𝑛+1 2𝑛+1 WB A3 Prove by induction that: LHS = RHS, true for n = 1 Assume that using assumption as required true for n = k + 1 if true for n = k. True for n = 1, true for all n.

n = 1 True for n = 1 n = k Assume True for n = k+1 WB A4 Prove by induction that: 1 𝑛 (2𝑟−1) 2 = 1 3 𝑛 4 𝑛 2 −1 𝐿𝐻𝑆 1 (2𝑟−1) 2 =2 1 −1=1 n = 1 𝑅𝐻𝑆 1 3 1 4 1 2 −1 =1 True for n = 1 1 𝑘 (2𝑟−1) 2 = 1 3 𝑘 4 𝑘 2 −1 n = k Assume True for n = k+1 true for n = k + 1 if true for n = k. True for n = 1, true for all n.

Replace the first part with the assumed formula from earlier! 𝑟=1 𝑛 𝑟2 𝑟 =2 1+(𝑛−1) 2 𝑛 WB A5 Prove by induction that: 𝐿𝐻𝑆 1 𝑟2 𝑟 =(1) 2 =2 n = 1 𝑅𝐻𝑆 2 1+(0)(2) =2 True for n = 1 𝑟=1 𝑘 𝑟2 𝑟 =2 1+(𝑘−1) 2 𝑘 n = k Assume 𝑟=1 𝑘+1 𝑟2 𝑟 = 2+8+24+64……+ 𝑘( 2 𝑘 )+ (𝑘+1) 2 𝑘+1 = 2 1+ 𝑘−1 2 𝑘 + (𝑘+1) 2 𝑘+1 Replace the first part with the assumed formula from earlier! =2+2(𝑘−1) 2 𝑘 + (𝑘+1) 2 𝑘+1 =2+(𝑘−1) 2 𝑘+1 + (𝑘+1) 2 𝑘+1 You need to aim for the power of 2 to be ‘k + 1’ (as it was ‘k’ originally) =2+(𝑘−1+𝑘+1) 2 𝑘+1 =2(1+𝑘 2 𝑘+1 ) True for n = k+1 NOW DO Ex 8A

Crucial points 1. Try to really understand the principle behind proof by induction Many students find this difficult, and it can take time for the ideas to really sink in. The Notes and Examples should help. 2. Always think about what you are aiming for When you take the assumed result for n = k and add on the (k + 1)th term, you want to rearrange this to get the formula for n = k + 1. It may help to actually write down the result you are looking for. 3. Be careful with algebraic manipulation In many questions on this topic, the algebraic manipulation required can appear daunting. It is vital that wherever possible you take out common factors. Never expand two or more terms and then try to factorise the result. See Example 1 in the Notes and Examples. 4. Make sure that you write out the proof correctly If you are proving a result using proof by induction, the tidying up of the result to prove the result for n = 1, or n = 2, as appropriate, is a vital part of the proof and

self-assess One thing learned is – One thing to improve is –

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Notes