OCR MEI Further Pure Core 2021 November — Question 7 6 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2021
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeGeometric series with summation
DifficultyChallenging +1.2 This is a proof by induction of a summation formula involving a product of linear and geometric terms. While it requires careful algebraic manipulation in the inductive step (particularly combining fractions with powers of 2), the structure is standard for Further Maths induction proofs. The formula is given, so no discovery is needed, and the technique is well-practiced at this level, making it moderately above average difficulty but not exceptional.
Spec4.01a Mathematical induction: construct proofs

7 Prove that \(\sum _ { r = 1 } ^ { n } \frac { r } { 2 ^ { r - 1 } } = 4 - \frac { n + 2 } { 2 ^ { n - 1 } }\) for all \(n \geqslant 1\).

Question 7:
AnswerMarks
71 rr 1 + 2
r = 1 = 4 − so true for n = 1
2 − 1 2 0
= 1
k rr k + 21
Assume true for n = k so r = 4 −
2 − 1 2 k −
= 1
k+1 r k+2 k+1
 =4− +
2r−1 2k−1 2k
r=1
2k+4−k−1
=4−
2k
k+1+2
=4− so true for n = k + 1
2k
So true for n = 1 and if true for n = k then true for
n = k + 1
AnswerMarks
 true for all nB1
M1
M1
M1
A1
A1
AnswerMarks
[6]2.1
2.1
2.1
1.1
2.2a
AnswerMarks
2.2aDependent on all previous marks
awarded
Question 7:
7 | 1 rr 1 + 2
r = 1 = 4 − so true for n = 1
2 − 1 2 0
= 1
k rr k + 21
Assume true for n = k so r = 4 −
2 − 1 2 k −
= 1
k+1 r k+2 k+1
 =4− +
2r−1 2k−1 2k
r=1
2k+4−k−1
=4−
2k
k+1+2
=4− so true for n = k + 1
2k
So true for n = 1 and if true for n = k then true for
n = k + 1
 true for all n | B1
M1
M1
M1
A1
A1
[6] | 2.1
2.1
2.1
1.1
2.2a
2.2a | Dependent on all previous marks
awarded
7 Prove that $\sum _ { r = 1 } ^ { n } \frac { r } { 2 ^ { r - 1 } } = 4 - \frac { n + 2 } { 2 ^ { n - 1 } }$ for all $n \geqslant 1$.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2021 Q7 [6]}}