OCR MEI C2 — Question 2 12 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeFind year when threshold exceeded
DifficultyModerate -0.3 This is a straightforward geometric sequences and series question from C2. Parts (i)-(ii) involve direct application of GP formulas (3^8 and sum of GP), part (iii) requires logarithm manipulation but follows a standard template for 'show that' inequalities, and part (iv) is a simple comparison calculation. While multi-part with 12 marks total, each component uses routine techniques with no novel problem-solving required, making it slightly easier than average.
Spec1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1
Jill has 3 daughters and no sons. They are generation 1 of Jill's descendants. Each of her daughters has 3 daughters and no sons. Jill's 9 granddaughters are generation 2 of her descendants. Each of her granddaughters has 3 daughters and no sons; they are descendant generation 3. Jill decides to investigate what would happen if this pattern continues, with each descendant having 3 daughters and no sons.
  1. How many of Jill's descendants would there be in generation 8? [2]
  2. How many of Jill's descendants would there be altogether in the first 15 generations? [3]
  3. After \(n\) generations, Jill would have over a million descendants altogether. Show that \(n\) satisfies the inequality $$n > \frac{\log_{10}2000003}{\log_{10}3} - 1.$$ Hence find the least possible value of \(n\). [4]
  4. How many fewer descendants would Jill have altogether in 15 generations if instead of having 3 daughters, she and each subsequent descendant has 2 daughters? [3]
Question 2:
AnswerMarks Guidance
2(i) 3×37 oe
6561M1
A1
AnswerMarks
[2]condone 1×37
or B2 if unsupporteddo not award if only seen in sum of
terms of GP
if 0, SC1 for 2187 unsupported
AnswerMarks Guidance
2(ii) valid attempt to sum a GP with r = 3
and n = 15
3(315 −1)
oe
3−1
AnswerMarks
21523359M1
M1
A1
AnswerMarks
[3]eg 3 + 32 + …….+ 315
or B2 if M1M0 or B3 if unsupportedmust see at least first two terms and
last term
NB 7174453 implies M1 from
1 + 3 + …+ 314
AnswerMarks Guidance
2(iii) 3(3n −1)
>1000000oe
3−1
2000000
eg 3n + 1 > 2000003 or 3n > +1
3
www
correctly taking logs of both sides
eg (n + 1) log3 > log2000003 or
nlog3 > log2000003 – log3
log2000003
egn+1> and completion to
log3
log2000003
n> −1
log3
AnswerMarks
n = 13 seenM1*
M1dep*
A1
B1
AnswerMarks
[4]eg log3n + 1 > log2000003 www
or log3n + log3 > log2000003 www;
may be implied by next stage of working
without any wrong working
AnswerMarks
B0 for n ≥ 13 or n > 13M0 for working backwards
M0 if = or < used
at least one previous progressive
interim step needed with no wrong
working;
M0dep* for log(3n – 1) > …
do not allow recovery from bracket
errors at any stage
AnswerMarks Guidance
2(iv) valid attempt to sum a GP with r = 2
and n = 15
their 21523359 – their 65534
AnswerMarks
21457825 iswM1*
M1dep*
A1
AnswerMarks
[3]if correct eg 2 + 22 + …….+ 215 = 65 534
with their 65534 < their 21523359
AnswerMarks
allow B3 for 21457825 unsupportedNB 32767 implies M1 from
1 + 2 +…+ 214
Question 2:
2 | (i) | 3×37 oe
6561 | M1
A1
[2] | condone 1×37
or B2 if unsupported | do not award if only seen in sum of
terms of GP
if 0, SC1 for 2187 unsupported
2 | (ii) | valid attempt to sum a GP with r = 3
and n = 15
3(315 −1)
oe
3−1
21523359 | M1
M1
A1
[3] | eg 3 + 32 + …….+ 315
or B2 if M1M0 or B3 if unsupported | must see at least first two terms and
last term
NB 7174453 implies M1 from
1 + 3 + …+ 314
2 | (iii) | 3(3n −1)
>1000000oe
3−1
2000000
eg 3n + 1 > 2000003 or 3n > +1
3
www
correctly taking logs of both sides
eg (n + 1) log3 > log2000003 or
nlog3 > log2000003 – log3
log2000003
egn+1> and completion to
log3
log2000003
n> −1
log3
n = 13 seen | M1*
M1dep*
A1
B1
[4] | eg log3n + 1 > log2000003 www
or log3n + log3 > log2000003 www;
may be implied by next stage of working
without any wrong working
B0 for n ≥ 13 or n > 13 | M0 for working backwards
M0 if = or < used
at least one previous progressive
interim step needed with no wrong
working;
M0dep* for log(3n – 1) > …
do not allow recovery from bracket
errors at any stage
2 | (iv) | valid attempt to sum a GP with r = 2
and n = 15
their 21523359 – their 65534
21457825 isw | M1*
M1dep*
A1
[3] | if correct eg 2 + 22 + …….+ 215 = 65 534
with their 65534 < their 21523359
allow B3 for 21457825 unsupported | NB 32767 implies M1 from
1 + 2 +…+ 214
Jill has 3 daughters and no sons. They are generation 1 of Jill's descendants.

Each of her daughters has 3 daughters and no sons. Jill's 9 granddaughters are generation 2 of her descendants. Each of her granddaughters has 3 daughters and no sons; they are descendant generation 3.

Jill decides to investigate what would happen if this pattern continues, with each descendant having 3 daughters and no sons.

\begin{enumerate}[label=(\roman*)]
\item How many of Jill's descendants would there be in generation 8? [2]
\item How many of Jill's descendants would there be altogether in the first 15 generations? [3]
\item After $n$ generations, Jill would have over a million descendants altogether. Show that $n$ satisfies the inequality
$$n > \frac{\log_{10}2000003}{\log_{10}3} - 1.$$
Hence find the least possible value of $n$. [4]
\item How many fewer descendants would Jill have altogether in 15 generations if instead of having 3 daughters, she and each subsequent descendant has 2 daughters? [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C2  Q2 [12]}}
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