Hint
The integer domain and starting value are stated. Use the claimed sum formula and show the substitution.
Worked solution
(a)
For \(\displaystyle n=1\) both sides equal 19. Assume the formula at \(\displaystyle n=k\). Then
\(\displaystyle S_{k+1}=\frac{10k(k+1)}2+9k+10(k+1)+9=\frac{10(k+1)(k+2)}2+9(k+1)\)
This is the required successor form. The base case and successor implication prove the formula for all integers \(\displaystyle n\ge 1\).
Answer: \(\displaystyle S_{n}= \frac{10n(n+ 1)}{2}+ 9n\) for all integers \(\displaystyle n\ge 1\).
(b)
\(10(12)+9=129\)
Answer: \(129\).
Answer: \(129\)
(c)
\(S_1=10+9=19\)
Answer: \(19\).
Answer: \(19\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 4 marks
- 1 mark: Verify the initial case on the specified integer domain.
- 1 mark: State and use the inductive assumption for an arbitrary k.
- 1 mark: Derive the required successor expression without assuming it.
- 1 mark: Conclude the claimed result for all integers in the stated domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Checking several numerical cases is not an inductive proof.
(b) · 2 marks
- 1 mark: Substitute n+1 in the summand, not in the whole sum.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not start by assuming the exact expression that must be proved for k+1.
(c) · 2 marks
- 1 mark: Evaluate both sides of the proposed identity at \(\displaystyle n=1\).
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Checking several numerical cases is not an inductive proof.
Total: 8 marks.