OCR MEI Further Pure Core 2020 November — Question 7 6 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2020
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve summation formula
DifficultyStandard +0.3 This is a straightforward proof by induction with a summation formula. The base case is trivial (n=1 gives 1×1! = 2!-1 = 1). The inductive step requires adding (n+1)×(n+1)! to both sides and factoring, which is a standard algebraic manipulation. While it's a Further Maths topic, the proof follows a completely standard template with no conceptual surprises, making it slightly easier than average overall.
Spec4.01a Mathematical induction: construct proofs
7 Prove by mathematical induction that \(\sum _ { r = 1 } ^ { n } ( r \times r ! ) = ( n + 1 ) ! - 1\) for all positive integers \(n\).
Question 7:
AnswerMarks
7When n = 1,
​ ​
Assume true for n = k so
​ ​ ​​
then
So if true for n = k then true for n = k+1
​ ​ ​​ ​ ​ ​​
As true for n = 1, true for all positive n
AnswerMarks
​ ​ ​B1
M1
M1
A1*
B1dep*
B1
AnswerMarks
[6]1.1
1.1
2.1
2.1
2.2a
AnswerMarks
2.4Or target seen
Final B mark only awarded if
all previous marks awarded
PPMMTT
Y420/01 Mark SchemeNovember 2020
8 (a) −λ = −1 + 2 , 2 + λ = 2 + 3 , 2 + 3λ = k + M1 3.1a
​ ​ µ​ ​ ​ ​ µ​ ​ ​ ​ ​​
4
µ​
λ = 3 − − M1 1.1
​ µ​ ​,​ ​3µ​ ​=​ ​1+2µ​
A1A1 1.1,1.1
⇒ = 1/5, λ = 3/5
µ​ ​ ​ ​ A1 1.1
2 + 9/5 = k + 4/5 ⇒ k = 3
​​ ​​ [5]
8 (b) DR
M1A1 1.1,1.1
B1 1.1 soi
⇒ θ = 43.3
​ ​ ​°
A1 1.1 accept 0.756 rad
[4]
9 (a)
B1 2.1 or y = mx + c
​​ ​ ​ ​
suppose y = mx is invariant
​​ ​ ​
M1 2.1 or λx + 3y = m(x – 2y)+c
λx + 3y = m(x – 2y) ​ ​ ​​ ​ ​ ​​ ​​ ​
​ ​​ ​ ​ ​​ ​​
⇒ λ + 3m = m(1− 2m)
​​ ​ ​ ​ ​ ​ ​
⇒ 2m2 ​+ 2m + λ = 0 A1 1.1
​ ​ ​ ​ ​​
no solutions if discriminant < 0 M1 3.1a
⇒ 4 − 8λ < 0, λ > ½ A1 3.2a
​​ ​​
[5]
9 (b) det M = 3 + 2λ or det M = −5 B1 1.1
​ ​ ​​ ​ ​
λ = −4 B1 2.1
​
m2 ​+ m – 2 = 0, m = −2 or 1
​ ​ ​ ​ ​
so lines are y = x and y = −2x B1 2.2a
​​ ​​ ​​ ​
[3]
10
Question 7:
7 | When n = 1,
​ ​
Assume true for n = k so
​ ​ ​​
then
So if true for n = k then true for n = k+1
​ ​ ​​ ​ ​ ​​
As true for n = 1, true for all positive n
​ ​ ​ | B1
M1
M1
A1*
B1dep*
B1
[6] | 1.1
1.1
2.1
2.1
2.2a
2.4 | Or target seen
Final B mark only awarded if
all previous marks awarded
PPMMTT
Y420/01 Mark SchemeNovember 2020
8 (a) −λ = −1 + 2 , 2 + λ = 2 + 3 , 2 + 3λ = k + M1 3.1a
​ ​ µ​ ​ ​ ​ µ​ ​ ​ ​ ​​
4
µ​
λ = 3 − − M1 1.1
​ µ​ ​,​ ​3µ​ ​=​ ​1+2µ​
A1A1 1.1,1.1
⇒ = 1/5, λ = 3/5
µ​ ​ ​ ​ A1 1.1
2 + 9/5 = k + 4/5 ⇒ k = 3
​​ ​​ [5]
8 (b) DR
M1A1 1.1,1.1
B1 1.1 soi
⇒ θ = 43.3
​ ​ ​°
A1 1.1 accept 0.756 rad
[4]
9 (a)
B1 2.1 or y = mx + c
​​ ​ ​ ​
suppose y = mx is invariant
​​ ​ ​
M1 2.1 or λx + 3y = m(x – 2y)+c
λx + 3y = m(x – 2y) ​ ​ ​​ ​ ​ ​​ ​​ ​
​ ​​ ​ ​ ​​ ​​
⇒ λ + 3m = m(1− 2m)
​​ ​ ​ ​ ​ ​ ​
⇒ 2m2 ​+ 2m + λ = 0 A1 1.1
​ ​ ​ ​ ​​
no solutions if discriminant < 0 M1 3.1a
⇒ 4 − 8λ < 0, λ > ½ A1 3.2a
​​ ​​
[5]
9 (b) det M = 3 + 2λ or det M = −5 B1 1.1
​ ​ ​​ ​ ​
λ = −4 B1 2.1
​
m2 ​+ m – 2 = 0, m = −2 or 1
​ ​ ​ ​ ​
so lines are y = x and y = −2x B1 2.2a
​​ ​​ ​​ ​
[3]
10
7 Prove by mathematical induction that $\sum _ { r = 1 } ^ { n } ( r \times r ! ) = ( n + 1 ) ! - 1$ for all positive integers $n$.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2020 Q7 [6]}}
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