Moderate -0.3 This is a standard proof by induction for divisibility, requiring routine application of the inductive hypothesis. The algebraic manipulation (factoring out 8 from 8^{k+1} - 3^{k+1}) is straightforward, and divisibility by 5 follows directly. Slightly easier than average since it's a textbook-style induction problem with no conceptual surprises.
=2(5𝑚)+5(cid:3435)8(cid:3038)(cid:3439)= 5(cid:3435)2𝑚+8(cid:3038)(cid:3439) div by 5
A1
so if true for n = k then true for n = k+1
Answer
Marks
Guidance
As true for n = 1, true for all n.
A1
must receive all previous marks for this to be awarded
[5]
Question 8:
8 | when n = 1, 8n – 3n = 5 div by 5
[Assume true when n = k] so 8k = 3k + 5m
8k+1 – 3k+1= 8(3k + 5m) – 33k
= 53k + 40m = 5(3k + 8m) div by 5
so if true for n = k then true for n = k+1
As true for n = 1, true for all n. | B1
M1
M1
A1
A1 | 2.1
2.1
2.1
2.2a
2.4 | or 3k = 8k – 5m or 8k − 3k = 5m or 8k − 3k div 5
assumption used
successful completion
3k = 8k – 5m used 5(8(cid:3038) +3𝑚) div 5 www
8k − 3k = 5m used 5(8𝑚+3(cid:3038)) div 5 www
must receive all previous marks for this to be awarded
Alternative method
when n = 1, 8n – 3n = 5 div by 5 | B1
[Assume true when n = k] so 𝑢 = 8k− 3k = 5m
(cid:3038) | M1
𝑢 −𝑢 = 8(cid:3038)(cid:2878)(cid:2869)− 3(cid:3038)(cid:2878)(cid:2869)− (8(cid:3038) − 3(cid:3038))
(cid:3038)(cid:2878)(cid:2869) (cid:3038) | M1 | considering difference between 𝑢 and 𝑢
(cid:3038)(cid:2878)(cid:2869) (cid:3038)
=2(cid:3435)8(cid:3038) −3(cid:3038)(cid:3439)+5(8(cid:3038))
=2(5𝑚)+5(cid:3435)8(cid:3038)(cid:3439)= 5(cid:3435)2𝑚+8(cid:3038)(cid:3439) div by 5 | A1
so if true for n = k then true for n = k+1
As true for n = 1, true for all n. | A1 | must receive all previous marks for this to be awarded
[5]
8 Prove by mathematical induction that $8 ^ { n } - 3 ^ { n }$ is divisible by 5 for all positive integers $n$.
\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q8 [5]}}