1. Proof from exam papers

2. MEI C1 Jan 2006 1. n is a positive integer. Show that n 2 + n is always even. [2]

3. Mark scheme

4. Examiners’ report 1) This proved to be a testing starter. Most candidates used a few different values, often some odd and some even – several commented that they had thereby proved the result by exhaustion! In the better solutions there was a realisation that generalisation was required, with good arguments being produced, the n 2 + n being more popular than the n(n + 1) form. A few good candidates substituted 2k and 2k + 1.

5. MEI C1 Jun 2009 6 Prove that, when n is an integer, n 3 − n is always even. [3]

6. Mark scheme

7. Examiners’ report 6) Many candidates tried only examples of odd and even numbers and did not attempt a general argument. Of those that did, the majority used the approach odd 3 = odd, odd −odd = even (and similarly with even numbers). Some factorised the expression as n(n 2 − 1) and argued successfully from there. Very few factorised it further. Some of the better candidates used n = 2m and n = 2m + 1, but some of these attempts were spoiled by errors in their algebra, particularly in expanding and simplifying (2m + 1) 3 − (2m + 1).

8. MEI C3 Jun 2009 7 (i) Show that (A) (x −y)(x 2 +xy +y 2 ) = x 3 −y 3, (B) (x + ½y) 2 + ¾y 2 = x 2 +xy +y 2. [4] (ii) Hence prove that, for all real numbers x and y, if x > y then x 3 > y 3. [3]

9. Mark scheme

10. Examiners’ report 7) (i) This algebra proved to be an easy 4 marks for all candidates, give or take a few slips due to carelessness. (ii) On the other hand, the logic of this part eluded all but the very best candidates. Many substituted values, or tried other letters, or y = x + 1, etc. Some recognised that x 2 + xy + y 2 had to be proved to be positive, but failed to see the connection between this and (i)(B).