Challenging +1.2 This is a proof by induction requiring students to verify the base case n=9 and handle the inductive step. While the algebraic manipulation in the inductive step (showing (n+1) > 4 for n ≥ 9) is straightforward, it's a Further Maths question requiring formal proof technique and some insight into how to structure the argument. The inequality itself is non-standard enough to be slightly above average difficulty.
So true for n = k ⇒ true for n = k + 1. But true for
Answer
Marks
n = 9. So true for all integers n ≥ 9
B1
M1
M1
A1
A1
Answer
Marks
[5]
2.1
2.1
1.1
2.2a
Answer
Marks
2.4
Basis case. Comparison must be
explicit and correct
Inductive hypothesis set up
Considering for k + 1 and using
inductive hypothesis correctly
Showing enough working to establish
statement for k + 1
Answer
Marks
Clear and complete conclusion
Bod sight of “true for n=1”
Might see 22k = 4k throughout
Allow M1 for use of inductive n = k
step when showing
Might see ≥ 10 or ≥ 𝑃𝑃9 k + fo 1 r →> 9𝑃𝑃. k
A0 here if k > 9 stated earlier
Do not allow if implication shown in
the direction
𝑃𝑃k+1 →𝑃𝑃k
Question 6:
6 | If n = 9, LHS = 9! = 362880
RHS = 218 = 262144 < LHS
[So true for n = 9]
Assume that k! > 22k for some k ≥ 9.
(k + 1)! = (k + 1)k! > (k + 1)×22k ...
... > 9×22k > 4×22k = 22×22k = 22 + 2k = 22(k+ 1)
ie (k + 1)! > 22(k+ 1)
So true for n = k ⇒ true for n = k + 1. But true for
n = 9. So true for all integers n ≥ 9 | B1
M1
M1
A1
A1
[5] | 2.1
2.1
1.1
2.2a
2.4 | Basis case. Comparison must be
explicit and correct
Inductive hypothesis set up
Considering for k + 1 and using
inductive hypothesis correctly
Showing enough working to establish
statement for k + 1
Clear and complete conclusion | Bod sight of “true for n=1”
Might see 22k = 4k throughout
Allow M1 for use of inductive n = k
step when showing
Might see ≥ 10 or ≥ 𝑃𝑃9 k + fo 1 r →> 9𝑃𝑃. k
A0 here if k > 9 stated earlier
Do not allow if implication shown in
the direction
𝑃𝑃k+1 →𝑃𝑃k