Proof from exam papers

MEI C1 Jan 2006

1. n is a positive integer. Show that n2 + n is always even. [2]

Mark scheme

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 n2 + n being more popular than the n(n + 1) form. A few good candidates substituted 2k and 2k + 1.

MEI C1 Jun 2009

6 Prove that, when n is an integer, n3 − n is always even. [3]

Mark scheme

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 odd3 = odd, odd −odd = even (and similarly with even numbers). Some factorised the expression as n(n2 − 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).

MEI C3 Jun 2009

7 (i) Show that(A) (x −y)(x 2+xy +y2) = x3 −y3,(B) (x + ½y)2 + ¾y2 = x2 +xy +y2 . [4] (ii) Hence prove that, for all real numbers x and

y, if x > y then x3 > y3. [3]

Mark scheme

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 x2 + xy + y2 had to be proved to be positive, but failed to see the connection between this and (i)(B).