Edexcel (9-1) Further Pure Math : Series - Problems and Solution

Sigma Notation and Properties

1. $\displaystyle\sum_{i=i_0}^{n} c a_i = c \sum_{i=i_0}^{n} a_i$ where $c$ is any number. So, we can factor constants out of a summation.

2. $\displaystyle\sum_{i=i_0}^{n} (a_i \pm b_i) = \sum_{i=i_0}^{n} a_i \pm \sum_{i=i_0}^{n} b_i$ So, we can break up a summation across a sum or difference.

3. $\displaystyle\sum_{i=1}^{n} c = cn$

4. $\displaystyle\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$

5. $\displaystyle\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$

6. $\displaystyle\sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2$

Problem 1

The third term of an arithmetic series is 70 and the sum of the first 10 terms of the series is 450.

• Calculate the common difference of the series.

The sum of the first $n$ terms of the series is $S_{n}$.

Given that $S_{n} \geqslant 350$

• Find the set of possible values of $n$.

Solution: Let the first term be $a$ and the common difference be $d$. The third term: $\d{a + 2d = 70} \quad\ldots (1)$ The sum of the first 10 terms: $S_{10} = 450$ $\d \frac{10}{2}\times (2a + 9d) = 450$ $2a + 9d = 90 \quad\ldots (2)$ From (1): $a = 70 - 2d$ Substitute $a$ in (2): $2(70 - 2d) + 9d = 90$ $140 - 4d + 9d = 90$ $5d = -50 \Rightarrow d = -10$ $\therefore\quad a = 70 - 2(-10) = 90$ For part (b), $\d S_n = \frac{n}{2} \times (2a + (n-1)d) \geqslant 350$ Substitute $a=90$ and $d = -10$: $\d{\frac{n}{2} \times (180 - 10n + 10) \geq 350}$ $\d{\frac{n}{2} \times (190 - 10n) \geq 350}$ $n(190 - 10n) \geq 700$ $190n - 10n^2 \geq 700$ $10n^2 - 190n + 700 \leq 0$ $n^2 - 19n + 70 \leq 0$ $(n-5)(n-14)\leq 0$ $5\leq n \leq 14$ $n\in\{5, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$

Problem 2

The sum of the first and third terms of a geometric series is 100.

The sum of the second and third terms is 60.

• Find the two possible values of the common ratio of the series.

Given that the series is convergent, find

• the first term of the series,
• the least number of terms for which the sum is greater than $159.9$.

Solution: Let the first term be $a$ and the common ratio be $r$. The first term: $a$ The second term: $ar$ The third term: $ar^2$ Given: $a + ar^2 = 100 \quad\ldots (1)$ $\d ar + ar^2 = 60 \quad\ldots (2)$ Dividing equation (1) by (2), $$\begin{aligned} & \frac{a+a r^2}{a r+a r^2}=\frac{100}{60} \\ & \frac{1+r^2}{r+r^2}=\frac{5}{3} \\ & 5 r+5 r^2=3+3 r^2 \\ & 2 r^2+5 r-3=0 \\ & (2 r-1)(r+3)=0 \\ & r=\frac{1}{2} \text { or } r=-3 \end{aligned}$$ Only $\d r = \frac{1}{2}$ is valid for a convergent series. For part (b), $$\begin{aligned} & a+a\left(\frac{1}{2}\right)^2=100 \\ & \frac{5}{4} a=100 \\ & a=80 \end{aligned}$$ For part (c), $\d S_n = \frac{a(1 - r^n)}{1 - r} > 159.9$: $\d S_n = \frac{80(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} > 159.9$ $\d 1-\left(\frac{1}{2}\right)^n>\frac{159.9}{160}$ $\d \left(\frac{1}{2}\right)^n < \frac{1}{1600}$ $\d \left(2\right)^n >1600$ $\d n >\log_2 1600\Rightarrow n=11$

Problem 3

The sum of the first and third terms of a geometric series $G$ is 104.

The sum of the second and third terms of $G$ is 24.

Given that $G$ is convergent and that the sum to infinity is $S$, find

• the common ratio of $G$.
• the value of $S$.

The sum of the first and third terms of another geometric series $H$ is also 104 and the sum of the second and third terms of $H$ is 24.

The sum of the first $n$ terms of $H$ is $S_{n}$.

• Write down the common ratio of $H$.
• Find the least value of $n$ for which $S_{n}>S$.

Solution: Let the first term be $a$ and the common ratio be $r$. The first term: $a$ The second term: $ar$ The third term: $ar^2$ $$\begin{aligned} & a+a r^2=104 \quad\ldots (1) \\ & a+a r=24 \quad\ldots (2) \\ & (1) \div(2) \\ & \frac{a+a r^2}{a+a r}=\frac{104}{24} \\ & \frac{1+r^2}{1+r}=\frac{13}{3} \\ & 3 r^2+3=13+13 r \\ & 3 r^2-13 r-10=0 \\ & (3 r+2)(r-5)=0 \\ & r=-\frac{2}{3} \text { or } r=5 \end{aligned}$$ Only $\d r = -\frac{21}{3}$ is valid for a convergent series. (b) Substituting $\d r=-\frac{2}{3}$ in equation (2), $$\begin{aligned} & a+a\left(-\frac{2}{3}\right)=2 y \\ & \frac{a}{3}=2 y \Rightarrow a=72 .\\ S & =\frac{72}{1-\left(-\frac{2}{3}\right)} \\ & =\frac{72}{\frac{5}{3}} \\ & =\frac{216}{5} \end{aligned}$$ (c) According to the part (a) the common ratio of $H$ is 5. (d) $$\begin{aligned} & a+5(a)=24 \\ & 6 a=24 \Rightarrow a=4 \\ & S_n=\frac{4\left(5^n-1\right)}{5-1}=5^n-1 \\ & S_n>S \\ & 5^n-1>\frac{216}{5} \\ & 5^n>\frac{221}{5} \\ & n>\log _5 \frac{221}{5} \\ & n>2.35 \\ & n=3 \end{aligned}$$

Problem 4

Evaluate $\d \sum_{n=6}^{20}(2n-3)$

Solution: \begin{aligned} \sum_{n=6}^{20}(2n-3) &= (2 \cdot 6 - 3) + (2 \cdot 7 - 3) + ... + (2 \cdot 20 - 3)\\ &= (12 - 3) + (14 - 3) + ... + (40 - 3)\\ &= 9 + 11 + 13 + ... + 37 \end {aligned} This is an arithmetic series with first term $9$, last term $37$, and common difference $2$. Number of terms: $\d n = \frac{37 - 9}{2} + 1 = 15$ Sum: $\d S = \frac{15}{2} \cdot (9 + 37) = 345$

Problem 5

The first four terms of an arithmetic series, $S$, are

$\log_{a} 2+\log_{a} 4+\log_{a} 8+\log_{a} 16$

• Write down an expression for the $r$ th term of $S$.
• Find an expression for the common difference of $S$.

The sum of the first $n$ terms of $S$ is $S_{n}$.

• Show that $S_{n}=\frac{1}{2} n(n+1) \log_{a} 2$.

The first four terms of a second arithmetic series, $T$, are

$\log_{a} 6+\log_{a} 12+\log_{a} 24+\log_{a} 48$

The sum of the first $n$ terms of $T$ is $T_{n}$.

• Find $T_{n}-S_{n}$ and simplify your answer.

Solution: (a) the $r$th term of $S$ is $\log_{a} (2^r) = r \log_{a} 2$. (b) the common difference of $S$ is $\log_{a} 4 - \log_{a} 2 = 2 \log_{a} 2 - \log_{a} 2 = \log_{a} 2$. (c) $$\begin{aligned} S_n & =\frac{n}{2}\left[2 \log _a 2+(n-1) \log _a 2\right] \\ & =\frac{n}{2}[2+n-1] \log _a 2 \\ & =\frac{1}{2} n(n+1) \log _a 2 \end{aligned}$$ (d) the first four terms of $T$ are $\log_{a} 6 + \log_{a} 12 + \log_{a} 24 + \log_{a} 48$. $$\begin{aligned} u_1 & =\log _a^b \\ u_2 & =\log _a 12=\log _a(6 \times 2) \\ u_3 & =\log _a 24=\log _a\left(6 \times 2^2\right) \\ u_4 & =\log _a 48=\log _a\left(6 \times 2^3\right) \\ u_r & =\log _a\left(6.2^{r-1}\right) \\ & =\log _a 6+\log _a 2^{r-1} \end{aligned}$$ $$\begin{aligned} T_n & =\sum_{r=1}^n u_r \\ & =\sum_{r=1}^n\left(\log _a^6+\log _a 2^{r-1}\right) \\ & =\sum_{r=1}^n \log _a 6+\sum_{r=1}^n \log _a 2^{r-1} \\ & =\sum_{r=1}^n \log _a 6+\sum_{r=1}^n(r-1) \log _a 2 \\ & =\sum_a^n \log _a 6+\sum_{r=1}^n r \log _a 2-\sum_{r=1}^n\log _a 2 \\ & =n \log _a 6+\frac{1}{2} n(n+1) \log _a 2-n \log _a 2\\ & =n \left(\log _a 6- \log _a 2\right)+\frac{1}{2} n(n+1) \log _a 2\\ & =n \log _a \frac{6}{2}+\frac{1}{2} n(n+1) \log _a 2\\ & =n \log _a 3+\frac{1}{2} n(n+1) \log _a 2 \end{aligned}$$ $\therefore\quad T_n-S_n=n \log _a 3$

Problem 6

The first term of a geometric series $S$ is $\sqrt{2}$.

The second term of $S$ is $\sqrt{2}-2$.

• (i) Find the exact value of the common ratio of $S$.

(ii) Find the third term of $S$, giving your answer in the form $a \sqrt{2}+b$, where $a$ and $b$ are integers.

• (i) Explain why the series is convergent.

(ii) Find the sum to infinity of $S$.

Solution: (a)(i) Let the first term be a and the common ratio be $r$ $$\begin{aligned} \therefore \quad a & =\sqrt{2} \\ a r & =\sqrt{2}-2 \\ \therefore \quad r & =\frac{\sqrt{2}-2}{\sqrt{2}}\\ \therefore\quad r&=1-\sqrt{2} \end{aligned}$$ (ii) $$\begin{aligned} \text { third term }& =ar^2 \\ & =(\sqrt{2}-2)(1-\sqrt{2}) \\ & =\sqrt{2}-2-2+2 \sqrt{2} \\ & =3 \sqrt{2}-4 \end{aligned}$$ (b) $$\begin{aligned} & r=1-\sqrt{2}=-0.414 \\ & |r|=0.414<1 \end{aligned}$$ (i) Hence the series a convergent. (ii) Sum to infinity $\d =\frac{\sqrt{2}}{1-(1-\sqrt{2})}=1$

Problem 7

• Show that $\d\sum_{r=1}^{n}(3 r-4)=\frac{n}{2}(3 n-5)$.
• Hence, or otherwise, evaluate $\d\sum_{r=11}^{50}(3 r-4)$.

Given that $\d \sum_{r=1}^{n}(3 r-4)=186$,

• find the value of $n$.

Solution: Method 1 $$\begin{aligned} \sum_{r=1}^n(3 r-4) & =(3-4)+(6-4)+(9-4)+\cdots+(3 n-4) \\ & =-1+2+5+\cdots+(3 n-4) \\ & =\frac{n}{2}(-1+3 n-4) \\ & =\frac{n}{2}(3 n-5) \end{aligned}$$ Method 2 $$\begin{aligned} \sum_{r=1}^n(3 r-4) & =\sum_{r=1}^n 3 r-\sum_{r=1}^n 4 \\ & =\frac{3 n}{2}(n+1)-4 n \\ & =\frac{n}{2}(3 n+3)-\frac{n}{2}(8) \\ & =\frac{n}{2}(3 n+3-8) \\ & =\frac{n}{2}(3 n-5) \end{aligned}$$ (b) $$\begin{aligned} \sum_{r=11}^{50}(3 r-4) & =\sum_{r=1}^{50}(3 r-4)-\sum_{r=1}^{10}(3 r-4) \\ & =\frac{50}{2}(3 \times 50-5)-\frac{10}{2}(3 \times 10-5) \\ & =25(145)-5(25) \\ & =25 \times 140 \\ & =3500 \end{aligned}$$ (c) $$\begin{aligned} & \sum_{r=1}^n(3 r-4)=186 \\ & \frac{n}{2}(3 n-5)=18-6 \\ & 3 n^2-5 n-372=0 \\ & (3 n+31)(n-12)=0\\ \therefore\quad & n=12\quad (\because n \text{ is a positive integer}) \end{aligned}$$

Problem 8

The sum $S_{n}$ of the first $n$ terms of an arithmetic series is given by $S_{n}=n(2 n+3)$. The first term of the series is $a$.

• Show that $a=5$.
• Find the common difference of the series.
• Find the 12th term of the series.

Given that $1+S_{p+4}=2 S_{p}$,

• find the value of $p$.

Solution: (a) $$\begin{aligned} S_n & =n(2 n+3) \\ 4 & =S_1=1(2 \times 1+3)=5 \\ \therefore\quad a & =5 \end{aligned}$$ (b) Let the common difference be $d$. $$\begin{aligned} & S_2=2(2 \times 2+3) \\ & u_1+u_2=14 \\ & a+a+d=14 \\ & 10+d=14 \\ & d=4 . \end{aligned}$$ (c) $u_{12}=a+11 d=49$ (d) $$\begin{aligned} & 1+S_{p+4}=2S_p \\ & 1+(p+4)(2(p+4)+3)=2 p(2 p+3) \\ & 1+(p+4)(2 p+11)=4 p^2+6 p \\ & 2 p^2+19 p+45=4 p^2+6 p \\ & 2 p^2-13 p-45=0 \\ & (2 p+5)(p-9)=0 \\ & p=-\frac{5}{2}, p=9 \end{aligned}$$ Since $p$ is a positive integer, The valid solution is $p=9$.

Problem 9

An arithmetic series has first term $a$ and common difference $d$. The $n$th term of the series is $t_{n}$ and the sum of the first $n$ terms of the series is $S_{n}$.

• Write down an expression in terms of $a$ and $d$ for

(i) $t_{58}$, (ii) $S_{13}$.

Given that $t_{58}=S_{13}$,

• show that $\d d=-\frac{4}{7} a$,
• show that $t_{176}=S_{21}$,
• find the value of $r$ when $t_{r}=5 t_{9}$.

Solution: (a) (i) $$\begin{aligned} t_{58} & =a+57 d \\ \text { (ii) } S_{13} & =\frac{13}{2}(2 a+12 d)=13 a+78 d \\ \text { (b) } t_{58} & =S_{13} \\ a+5-7 d & =13 a+78 d \\ 21 d & =-12 a \\ d & =-\frac{4}{7} a \end{aligned}$$ (c) $$\begin{aligned} t_{176} & =a+175 d \\ & =a+175\left(-\frac{4}{7} a\right) \\ & =-99 a\\ S_{21} & =\frac{21}{2}(2 a+20 d) \\ & =21\left[a+10\left(-\frac{4}{7} a\right)\right] \\ & =21 a-120 a \\ & =-99 a \\ \therefore \quad t_{176} & =S_{21} \end{aligned}$$ (d) $$\begin{aligned} & t_r=5 t_9 \\ & a+(r-1) d=5(a+8 d) \\ & a+(r-1)\left(-\frac{4}{7} a\right)=5\left(a+8\left(-\frac{4}{7} a\right)\right) \\ & 1-\frac{4 r}{7}+\frac{4}{7}=5-\frac{160}{7} \\ & r=34 \end{aligned}$$

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