Proof – Year 12 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Odd numbers have the form \(2k+1\); consecutive odd numbers differ by 2. \(2n-1\) and \(2n+1\) (1)
   2. Expand: \((2n-1)^2=4n^2-4n+1\) and \((2n+1)^2=4n^2+4n+1\); the \(\pm4n\) terms cancel. \((2n-1)^2+(2n+1)^2=8n^2+2\) (2). This is \(8n^2+2\), which is 2 more than the multiple of 8 given by \(8n^2\) (1). Since \(n^2\) is an integer, the sum is always 2 more than a multiple of 8.

   Using \(n\) and \(n+2\) is wrong because they need not be odd.

2. Answer to Question 2 (4 marks)

   1. One counter-example is enough to disprove a general statement; try small primes. \(n=7\): \(7+2=9\) is not prime (2) So the statement is false.
   2. Choose simple non-zero values of \(a\) and \(b\). \(a=1\), \(b=2\): \((1+2)^2=9\) but \(1^2+2^2=5\) (2) So \((a+b)^2\ne a^2+b^2\) in general; the correct expansion is \(a^2+2ab+b^2\).

   The marks require the calculation as well as the example. In (a), \(n=2\) also works, since \(2+2=4\) is not prime.

3. Answer to Question 3 (4 marks)

   Every integer is either even or odd, so checking these two cases covers all integers. Write each case algebraically and factorise out 2.

   If \(n=2k\): \(n^2+n=4k^2+2k=2(2k^2+k)\), even (2). If \(n=2k+1\): \((2k+1)^2+(2k+1)=4k^2+6k+2=2(2k^2+3k+1)\), even (2). Both cases are even so the statement is true.

   Alternatively, \(n^2+n=n(n+1)\) is a product of consecutive integers, one of which must be even.

   Checking a few values such as 2 and 3 is not a proof.

4. Answer to Question 4 (4 marks)

   In a proof by contradiction, assume the opposite of the statement and show that it leads to something impossible.

   Assume there is a largest even number \(N\) (1). Then \(N+2\) is also even (1) and \(N+2\gt N\) (1), which contradicts the assumption. So there is no largest even number (1).

   The same argument shows that there is no largest integer of any kind.

5. Answer to Question 5 (4 marks)

   1. The numbers of seats form an arithmetic sequence with first term 10 and common difference 2. \(10+2(n-1)=2n+8\) (1) This is the required expression.
   2. Factorise out 2. \(2n+8=2(n+4)\) (1) Since \(n+4\) is an integer, every row has an even number of seats.
   3. Use \(S_n=\tfrac n2[2a+(n-1)d]\) with \(n=15\), \(a=10\), \(d=2\). Arithmetic series: \(\tfrac{15}2(2\times10+14\times2)=\tfrac{15}{2}\times48=360\) (1 for the formula, 1 for 360) The hall has 360 seats in total.

Total : 20 marks