Hint
The recurrence and starting index must both be specified. Shift the recurrence by its fixed point or apply the closed form.
Worked solution
(a)
At \(\displaystyle n=1\) the expression equals 10. Assuming it at k, the recurrence gives \(u_{k+1}=2[(21)2^{k-1}+(-11)]+11=(21)2^k+(-11)\) because \(2L+11=L\). This proves the successor, so the formula holds for every \(\displaystyle n\ge 1\).
Answer: \(u_n=(21)2^{n-1}+(-11)\)
(b)
\(L=2L+11\implies L=-11\)
Answer: \(-11\).
Answer: \(-11\)
(c)
\(u_{5}=2u_{4}+11=325\)
Answer: \(325\).
Answer: \(325\)
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: A matching first few terms do not establish a general formula.
(b) · 2 marks
- 1 mark: Solve the constant-value recurrence equation.
- 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: Use \(\displaystyle r^{n−1}\) for a sequence that starts at \(\displaystyle u_{1}\).
(c) · 2 marks
- 1 mark: Use the recurrence once after the requested term.
- 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: A matching first few terms do not establish a general formula.
Total: 8 marks.