Proof by induction – further maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   \(n=1\): \(1\times3=3\) and \(\tfrac16\times1\times2\times9=3\) ✓ [1]. Assume \(\displaystyle\sum_{r=1}^{k}r(r+2)=\tfrac16k(k+1)(2k+7)\) [1]. Then the sum to \(k+1\) is \(\tfrac16k(k+1)(2k+7)+(k+1)(k+3)=\tfrac16(k+1)(2k^2+13k+18)=\tfrac16(k+1)(k+2)(2k+9)\) [1], which is the formula for \(n=k+1\). Since the result is true for \(n=1\), and if it is true for \(n=k\) then it is true for \(n=k+1\), by mathematical induction it is true for all positive integers \(n\). [1]

   The key algebra, after taking out \(\tfrac16(k+1)\): \(k(2k+7)+6(k+3)=2k^2+13k+18=(k+2)(2k+9)\).

2. Answer to Question 2 (4 marks)

   \(f(1)=10=5\times2\) ✓ [1]. Assume \(6^k+4=5m\) [1]. Then \(6^{k+1}+4=6\cdot6^k+4=6(5m-4)+4=30m-20\) [1] \(=5(6m-4)\), divisible by 5. Since the result is true for \(n=1\), and if it is true for \(n=k\) then it is true for \(n=k+1\), by mathematical induction it is true for all positive integers \(n\). [1]

   An alternative inductive step uses \(f(k+1)-f(k)=6^{k+1}-6^k=5\times6^k\), a multiple of 5, so \(f(k+1)=f(k)+5\times6^k\) is also a multiple of 5. Check \(n=2\): \(36+4=40\) ✓.

3. Answer to Question 3 (4 marks)

   \(n=1\): \(\begin{pmatrix}3&-4\\1&-1\end{pmatrix}\) ✓ [1]. Assume true for \(n=k\). \(\begin{pmatrix}1+2k&-4k\\k&1-2k\end{pmatrix}\begin{pmatrix}3&-4\\1&-1\end{pmatrix}=\begin{pmatrix}3+6k-4k&-4-8k+4k\\3k+1-2k&-4k-1+2k\end{pmatrix}\) [1] \(=\begin{pmatrix}3+2k&-4-4k\\k+1&-1-2k\end{pmatrix}=\begin{pmatrix}1+2(k+1)&-4(k+1)\\k+1&1-2(k+1)\end{pmatrix}\) [1]. Since the result is true for \(n=1\), and if it is true for \(n=k\) then it is true for \(n=k+1\), by mathematical induction it is true for all positive integers \(n\). [1]

   Check \(n=2\): the formula gives \(\begin{pmatrix}5&-8\\2&-3\end{pmatrix}\), and squaring the matrix directly gives \(9-4=5\), \(-12+4=-8\), \(3-1=2\) and \(-4+1=-3\) ✓. Multiplying on the left instead of on the right is also valid here.

4. Answer to Question 4 (4 marks)

   1. \(u_2=17\), \(u_3=65\) [1]
   2. \(u_1=4+1=5\) ✓ [1]. Assume \(u_k=4^k+1\). Then \(u_{k+1}=4(4^k+1)-3=4^{k+1}+1\) [1]. Since the result is true for \(n=1\), and if it is true for \(n=k\) then it is true for \(n=k+1\), by mathematical induction it is true for all positive integers \(n\). [1]

   In (a), \(u_2=4\times5-3=17\) and \(u_3=4\times17-3=65\), which match \(4^2+1\) and \(4^3+1\). The key step is \(4\times4^k+4-3=4^{k+1}+1\), where the assumption is used explicitly.

5. Answer to Question 5 (4 marks)

   1. \(2^4=16=4^2\); \(2^5=32\gt 25\) ✓ [1]
   2. \(n=5\): \(32\gt 25\) ✓. Assume \(2^k\gt k^2\) for some \(k\ge5\) [1]. Then \(2^{k+1}=2\times2^k\gt 2k^2\) [1] \(\gt (k+1)^2\). So true for \(n=k+1\), and by induction true for all \(n\ge5\) [1].

   The base case is \(n=5\), not \(n=1\): the statement fails for \(n=2,3,4\), since for example \(2^3=8\lt9\). The chain \(2^{k+1}\gt2k^2\gt(k+1)^2\) uses the assumption first and the given fact second. Check the given fact at \(k=5\): \(50\gt36\) ✓. After 6 weeks, A has 64 followers and B has 36.

Total : 20 marks