Standard +0.3 This is a straightforward induction proof with a simple divisibility statement. The algebra is routine (expanding (n+1)³ + (n+2)³ + (n+3)³ and showing it equals the previous sum plus a multiple of 9), and the structure follows a standard template. Slightly above average difficulty only because it's Further Maths and requires careful algebraic manipulation, but no novel insight is needed.
But it is true for r =1 so is true for all integers r
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Answer
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2.1
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Basis case – either 4×9 or “36 is divisible by
9” given explicitly.
For getting started (Sight of f(r) is not
necessary)
For finding f(r + 1)
Each term as a multiple of 9
Conclusion dependant on all other marks
being earned
[5]
Question 7:
7 | Basis case: 13+23+33 =36=4×9
Consider the sum f(r)=r3+ ( r+1 )3 + ( r+2 )3
Assume that f(r)=9k for some k∈Ζ
Then f(r+1)=f(r)+ ( r+3 )3 −r3
=f(r)+r3 +3r2×3+3r×9+27 −r3
=f(r)+9r2 +27r+27
( )
= 9k+9 r2 +3r+3 =9k' for some k'∈Ζ
So if true for r then true also for r+1
But it is true for r =1 so is true for all integers r | B1
M1
M1
A1
A1 | 2.1
2.1
2.1
2.2a
2.5 | Basis case – either 4×9 or “36 is divisible by
9” given explicitly.
For getting started (Sight of f(r) is not
necessary)
For finding f(r + 1)
Each term as a multiple of 9
Conclusion dependant on all other marks
being earned
[5]
7 Prove by induction that the sum of the cubes of three consecutive positive integers is divisible by 9 .
\hfill \mbox{\textit{OCR Further Pure Core 1 2020 Q7 [5]}}