Proof

Prove by induction that for all positive integers n, \sum_{r=1}^{n} r(r+1) = \frac{n(n+1)(n+2)}{3}.",
"marks": 6,
"solution": "Step 1: Base Case (n=1)
LHS = 1(1+1) = 2
RHS = 1(1+1)(1+2)/3 = 6/3 = 2
LHS = RHS, so the statement is true for n=1.

Step 2: Inductive Hypothesis
Assume the statement is true for n=k, so \sum_{r=1}^{k} r(r+1) = \frac{k(k+1)(k+2)}{3}.

Step 3: Inductive Step (n=k+1)
We need to prove that \sum_{r=1}^{k+1} r(r+1) = \frac{(k+1)(k+2)(k+3)}{3}.
\sum_{r=1}^{k+1} r(r+1) = \sum_{r=1}^{k} r(r+1) + (k+1)(k+2)
Using the assumption: = \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)
Factor out (k+1)(k+2): = (k+1)(k+2) [\frac{k}{3} + 1]
= (k+1)(k+2) [\frac{k+3}{3}]
= \frac{(k+1)(k+2)(k+3)}{3}
This is the required result for n=k+1.

Step 4: Conclusion
Since the statement is true for n=1, and if it is true for n=k then it is true for n=k+1, by the principle of mathematical induction, the statement is true for all positive integers n.",
"examiner_commentary": "This response would score full marks. It clearly shows the base case (B1), states the assumption (M1), and uses it correctly in the inductive step (M1). The algebraic manipulation to factorise and reach the target expression is clear and correct (A1). The final concluding statement is essential and secures the final A1 cso mark. A common error is to expand the expression instead of factorising, which is more prone to algebraic mistakes.

Prove by induction that f(n) = 7^n - 1 is divisible by 6 for all integers n \ge 1.",
"marks": 5,
"solution": "Step 1: Base Case (n=1)
f(1) = 7^1 - 1 = 6. 6 is divisible by 6, so the statement is true for n=1.

Step 2: Inductive Hypothesis
Assume the statement is true for n=k, so f(k) = 7^k - 1 is divisible by 6. This means we can write 7^k - 1 = 6A for some integer A.

Step 3: Inductive Step (n=k+1)
We need to prove that f(k+1) = 7^{k+1} - 1 is divisible by 6.
f(k+1) = 7^{k+1} - 1 = 7 \cdot 7^k - 1
From the assumption, 7^k = 6A + 1. Substitute this in:
f(k+1) = 7(6A + 1) - 1
= 42A + 7 - 1
= 42A + 6
= 6(7A + 1)
Since A is an integer, 7A+1 is also an integer. Therefore, f(k+1) is a multiple of 6.

Step 4: Conclusion
Since the statement is true for n=1, and if it is true for n=k then it is true for n=k+1, by the principle of mathematical induction, the statement is true for all integers n \ge 1.",
"examiner_commentary": "This is a full-mark solution. Credit is given for the base case (B1), the assumption with the definition of an integer multiplier (M1), and the correct substitution in the inductive step (M1, A1). The alternative method of considering f(k+1) - f(k) also works well here. A frequent mistake is to write 'f(k) is divisible by 6' but not to express it algebraically as f(k) = 6A, which is needed for the substitution.

Let the matrix M = \begin{pmatrix} 3 & -2 \ 1 & 0 \end{pmatrix}. Prove by induction that for n \ge 1, M^n = \begin{pmatrix} 2^{n+1}-1 & 2-2^{n+1} \ 2^n-1 & 2-2^n \end{pmatrix}.",
"marks": 6,
"solution": "Step 1: Base Case (n=1)
LHS = M^1 = \begin{pmatrix} 3 & -2 \ 1 & 0 \end{pmatrix}
RHS = \begin{pmatrix} 2^{2}-1 & 2-2^{2} \ 2^1-1 & 2-2^1 \end{pmatrix} = \begin{pmatrix} 3 & -2 \ 1 & 0 \end{pmatrix}
LHS = RHS, so the statement is true for n=1.

Step 2: Inductive Hypothesis
Assume the statement is true for n=k, so M^k = \begin{pmatrix} 2^{k+1}-1 & 2-2^{k+1} \ 2^k-1 & 2-2^k \end{pmatrix}.

Step 3: Inductive Step (n=k+1)
We need to prove M^{k+1} = \begin{pmatrix} 2^{k+2}-1 & 2-2^{k+2} \ 2^{k+1}-1 & 2-2^{k+1} \end{pmatrix}.
M^{k+1} = M^k imes M = \begin{pmatrix} 2^{k+1}-1 & 2-2^{k+1} \ 2^k-1 & 2-2^k \end{pmatrix} \begin{pmatrix} 3 & -2 \ 1 & 0 \end{pmatrix}
= \begin{pmatrix} 3(2^{k+1}-1) + 1(2-2^{k+1}) & -2(2^{k+1}-1) + 0 \ 3(2^k-1) + 1(2-2^k) & -2(2^k-1) + 0 \end{pmatrix}
= \begin{pmatrix} 3 \cdot 2^{k+1}-3 + 2-2^{k+1} & -2 \cdot 2^{k+1}+2 \ 3 \cdot 2^k-3 + 2-2^k & -2 \cdot 2^k+2 \end{pmatrix}
= \begin{pmatrix} 2 \cdot 2^{k+1}-1 & 2-2^{k+2} \ 2 \cdot 2^k-1 & 2-2^{k+1} \end{pmatrix}
= \begin{pmatrix} 2^{k+2}-1 & 2-2^{k+2} \ 2^{k+1}-1 & 2-2^{k+1} \end{pmatrix}
This is the required result for n=k+1.

Step 4: Conclusion
Since the statement is true for n=1, and if it is true for n=k then it is true for n=k+1, by the principle of mathematical induction, the statement is true for all positive integers n.",
"examiner_commentary": "Full marks. The base case is clearly shown (B1). The assumption is stated (M1). The matrix multiplication M^k imes M is explicitly calculated, which is crucial for method marks (M1, A1). The algebraic simplification involving indices is correct (A1). The final conclusion is present and correct (A1). Candidates often lose marks by making small algebraic slips with the powers of 2 during the simplification.