A-LevelFurther MathematicsProof by induction9231/12 · May/June 2018 · Q26 Marks
It is given that f(n) = 2³ⁿ + 8ⁿ⁻¹. By simplifying f(k) + f(k + 1), or otherwise, prove by mathematical induction that f(n) is divisible by 9 for every positive integer n.
- It is given that f(n) = 2³ⁿ + 8ⁿ⁻¹. By simplifying f(k) + f(k + 1), or otherwise, prove by mathematical induction that f(n) is divisible by 9 for every positive integer n.[6]
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- B1 Checks base case. We have that f(1) = 9 is divisible by 9.
- B1 Makes general statement. Assume that f(k) is divisible by 9.
- M1 Uses expansion of f(k+1). f(k+1)+f(k) = 2^(3k+3) +8^k + 2^(3k) + 8^(k-1)
- A1 Correct split of powers. = 2^3.2^(3k)+8.8^(k-1)+ 2^(3k) + 8^(k-1) OE
- A1 = 9(2^(3k) +8^(k-1)) OE so f(k + 1) is divisible by 9.
- A1 So if f(k) is divisible by 9, so is f(k+1), (and f(1) is divisible by 9), f(n) is divisible by 9 for every integer n ≥1
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