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A-Level Pure 2 · Edexcel IAL WMA12 & CAIE 9709 · Topic pack

Topic pack · Geometric Series & Binomial แพ็กฝึกเฉพาะหัวข้อ · อนุกรมเรขาคณิต · ทวินาม

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Geometric series & binomial expansion, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

A geometric sequence has first term $64$ and common ratio $\dfrac12$. Find the 8th term of the sequence.

Mark schemeshow ▾
M1 $u_n=ar^{\,n-1}$  M1 $u_8=64\times\left(\dfrac12\right)^{7}=\dfrac{64}{128}$  A1 $=\dfrac12$.
Q2. 4 marks

A geometric series begins $3+6+12+24+\cdots$. (a) Write down the common ratio. (b) Find the sum of the first 12 terms.

Mark schemeshow ▾
(a) B1 $r=2$.
(b) M1 $S_n=\dfrac{a(r^n-1)}{r-1}$  M1 $S_{12}=\dfrac{3(2^{12}-1)}{2-1}=3\times4095$  A1 $=12285$.
Q3. 4 marks

Find the coefficient of $x^3$ in the binomial expansion of $(2+x)^7$.

Mark schemeshow ▾
M1 relevant term $\dbinom{7}{3}2^{4}x^{3}$  B1 $\dbinom{7}{3}=35$  M1 $35\times2^4=35\times16$  A1 coefficient $=560$.
Q4. 5 marks

The second term of a geometric series is $6$ and its sum to infinity is $27$. Find the two possible values of the common ratio $r$, and the corresponding first terms.

Mark schemeshow ▾
M1 $ar=6$ and $\dfrac{a}{1-r}=27$  M1 eliminate $a$: $27r(1-r)=6$  A1 $9r^2-9r+2=0$, $(3r-1)(3r-2)=0$, so $r=\dfrac13$ or $r=\dfrac23$ (both valid, $|r|<1$)  M1 $a=27(1-r)$  A1 $r=\dfrac13,\ a=18$; $r=\dfrac23,\ a=9$.
Q5. 5 marks

(a) Find the first four terms, in ascending powers of $x$, of the binomial expansion of $(1-2x)^5$. (b) By substituting $x=0.01$, use your expansion to estimate $0.98^5$, giving your answer to 5 decimal places.

Mark schemeshow ▾
(a) M1 $\dbinom{5}{k}(-2x)^k$ terms  A1 $1-10x+40x^2$  A1 $-80x^3$.
(b) B1 substitute $x=0.01$: $1-0.1+0.004-0.00008$  A1 $0.98^5\approx0.90392$.
Q6. 6 marks

A machine is bought for £20 000. Its value depreciates by $15\%$ each year, so after $n$ years its value is £$20000\times0.85^n$. (a) Find the value of the machine after 3 years. (b) Find the number of complete years after which the value of the machine first falls below £8000.

Mark schemeshow ▾
(a) M1 $20000\times0.85^3$  A1 £$12282.50$.
(b) M1 $20000\times0.85^n<8000$, i.e. $0.85^n<0.4$  M1 take logs: $n\log0.85<\log0.4$ (inequality reverses, $\log 0.85<0$)  A1 $n>\dfrac{\log0.4}{\log0.85}=5.63\ldots$  A1 $n=6$ years.
Q7. 6 marks

The second term of a geometric series is $12$ and the fifth term is $324$. (a) Find the common ratio and the first term. (b) Find the least value of $n$ for which the sum of the first $n$ terms exceeds $100 000$.

Mark schemeshow ▾
(a) M1 $\dfrac{ar^4}{ar}=r^3=\dfrac{324}{12}=27$  A1 $r=3$  A1 $a=\dfrac{12}{3}=4$.
(b) M1 $S_n=\dfrac{4(3^n-1)}{2}=2(3^n-1)>100000$, so $3^n>50001$  M1 $n>\dfrac{\log50001}{\log3}=9.84\ldots$  A1 $n=10$ (check: $S_9=39364$, $S_{10}=118096$).
Q8. 7 marks

In the binomial expansion of $(1+ax)^8$, where $a>0$ is a constant, the coefficient of $x^2$ is $112$. (a) Find the value of $a$. (b) Find the coefficient of $x^3$ in the expansion. (c) Hence find the coefficient of $x^3$ in the expansion of $(3-x)(1+ax)^8$.

Mark schemeshow ▾
(a) M1 coefficient of $x^2$ is $\dbinom{8}{2}a^2=28a^2$  M1 $28a^2=112$, $a^2=4$  A1 $a=2$ (since $a>0$).
(b) M1 $\dbinom{8}{3}2^3=56\times8$  A1 $=448$.
(c) M1 $3\times448-1\times112$ (from $3\cdot x^3$ term and $-x\cdot x^2$ term)  A1 $=1232$.
Q9. 7 marks

Nan saves into a scheme. She pays in £100 in the first month, and each month she pays in $5\%$ more than the month before, so her payments form a geometric sequence. (a) Find the amount she pays in the 12th month, to the nearest penny. (b) Find the total she pays in over the first 24 months, to the nearest penny. (c) Find the number of the first month in which her payment exceeds £300.

Mark schemeshow ▾
(a) M1 $100\times1.05^{11}$  A1 £$171.03$.
(b) M1 $S_{24}=\dfrac{100(1.05^{24}-1)}{0.05}$  A1 £$4450.20$.
(c) M1 $100\times1.05^{\,n-1}>300$, i.e. $1.05^{\,n-1}>3$  M1 $n-1>\dfrac{\log3}{\log1.05}=22.5\ldots$  A1 $n-1=23$, so month $n=24$.
Q10. 8 marks

A geometric series has first term $a$ and common ratio $r$. The sum of its first two terms is $15$ and its sum to infinity is $27$. (a) Show that $r^2=\dfrac49$. (b) Find the two possible pairs of values of $a$ and $r$. (c) For the series with positive common ratio, find the least value of $n$ for which $S_\infty-S_n<0.5$.

Mark schemeshow ▾
(a) M1 $a(1+r)=15$ and $\dfrac{a}{1-r}=27$  M1 substitute $a=27(1-r)$: $27(1-r)(1+r)=15$  A1 $1-r^2=\dfrac{5}{9}$, so $r^2=\dfrac49$ ∎.
(b) A1 $r=\dfrac23$ or $r=-\dfrac23$  M1 $a=27(1-r)$  A1 $r=\dfrac23,\ a=9$; $r=-\dfrac23,\ a=45$.
(c) M1 $S_\infty-S_n=\dfrac{ar^n}{1-r}=27\left(\dfrac23\right)^n<0.5$, so $\left(\dfrac23\right)^n<\dfrac{1}{54}$; logs: $n>\dfrac{\log54}{\log(3/2)}=9.83\ldots$  A1 $n=10$.