Hint
There are n−1 multiplications in \(\displaystyle u_{n}\) but a power of n in the finite-sum formula.
Worked solution
(a)
\(u_8=3\cdot2^7=384\).
Answer: 384.
(b)
\(\displaystyle S_{8}= \frac{3(2^{8}- 1)}{2- 1}= 765\).
Answer: 765.
(c)
\(3(2^n-1)>3000\) gives \(2^n>1001\). Since \(\displaystyle 2^{9}=512\) and \(\displaystyle 2^{10}=1024\), the least index is 10. In fact \(\displaystyle S_{9}=1533\) and \(\displaystyle S_{10}=3069\); the positive terms make the partial sums increase.
Answer: \(n=10\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Use the power 7 for the eighth term.
- 1 mark: Evaluate 384.
Alternative methods: Repeated multiplication for eight correctly indexed terms is accepted.
Common error: Using \(\displaystyle 2^{8}\) in the eighth term.
(b) · 2 marks
- 1 mark: Use the finite GP sum with eight terms.
- 1 mark: Evaluate 765.
Alternative methods: Direct addition of the eight terms is accepted.
Common error: Trying to use an infinite sum despite ratio 2.
(c) · 3 marks
- 1 mark: Form \(\displaystyle 2^{n}\)>1001 or an equivalent correct sum inequality.
- 1 mark: Identify \(\displaystyle n=10\).
- 1 mark: Justify minimality with the neighbouring powers/totals and increasing sums.
Alternative methods: A logarithm calculation followed by the correct integer check is accepted.
Common error: Solving a term threshold rather than a total threshold.
Total: 7 marks.