8.1 Sequences and Series
A sequence is a list of numbers in a definite order, given by a formula for its $n$th term or by a rule that builds each term from earlier ones. Adding its terms gives a series, written compactly with $\Sigma$.
A doctor records a patient's temperature every hour: $38.9,\ 38.6,\ 38.2,\ 37.9,\dots$ . A cricket scorer notes the total runs at the end of each over. A bank statement lists your balance month by month. In each case the order matters โ the third reading comes after the second โ and often there is a rule: the table of $7$ is $7,14,21,28,\dots$, the squares are $1,4,9,16,\dots$, and a ball that bounces to three-quarters of its previous height reaches $12,\ 9,\ 6.75,\dots$ metres. Lists like these are called sequences.
Very often we want the total rather than the list โ the total runs, the total distance travelled by the ball, the total money saved. Adding the terms of a sequence gives a series. In this section we make both ideas precise, describe sequences by a formula for the $n$th term or by a rule that uses earlier terms (the Fibonacci numbers are the famous example), introduce the sigma notation $\Sigma$ for sums, and recall the arithmetic progression you studied in Class 10 โ the simplest sequence of all, and the model for the geometric progression of Section 8.2.
Sequences and Their Terms
A sequence is a list of numbers arranged in a definite order: a first number $a_1$, a second number $a_2$, a third number $a_3$, and so on. The numbers are the terms of the sequence, and $a_n$ (the term in position $n$) is called the $n$th term or general term.
Formally, a sequence is a function whose domain is the set of natural numbers $\mathbb{N}=\{1,2,3,\dots\}$ โ an infinite sequence $a_1,a_2,a_3,\dots$ โ or a set $\{1,2,\dots,n\}$ โ a finite sequence $a_1,a_2,\dots,a_n$ with exactly $n$ terms. The value of the function at $n$ is the term $a_n$.
So the sequence of even numbers is the function $n\mapsto 2n$: $a_1=2$, $a_2=4$, $a_{10}=20$. The digits of your mobile number form a finite sequence of $10$ terms; the natural numbers $1,2,3,\dots$ form an infinite one. Because a sequence is a function of $n$, we can draw it: plot the points $(n,a_n)$. The graph is a row of separate dots, one above each of $n=1,2,3,\dots$ โ never a joined-up curve, because there is no "term number $2.5$".
When we know a formula for $a_n$ in terms of $n$, we can find any term at once, without working out the ones before it. Not every sequence has such a formula: the prime numbers $2,3,5,7,11,13,\dots$ form a perfectly good sequence (the rule "list the primes in increasing order" decides every term), but no simple formula for the $n$th prime is known.
(b) Write the first five terms of $a_n=(-1)^{n+1}\,3n$.
(c) For $a_n=n^2-5n+3$, find $a_{12}$, and find which term equals $17$.
(a) $5,\ 9,\ 13,\ 17,\dots$ (b) $\dfrac12,\ \dfrac23,\ \dfrac34,\ \dfrac45,\dots$ (c) $2,\ -4,\ 8,\ -16,\dots$ (d) $0,\ 3,\ 8,\ 15,\ 24,\dots$
"Find the next term of $1,2,4,\dots$" has no single correct answer. Doubling gives $1,2,4,8,\dots$; but the rule "add $1$, then $2$, then $3$, โฆ" gives $1,2,4,7,11,\dots$, and both fit the three terms shown. In exams the intended pattern is usually the simplest one, but a sequence is only truly known when its rule is given โ a formula for $a_n$, or a recursive rule as in the next subsection.
Recursive Rules and the Fibonacci Sequence
Many sequences are described not by a formula for $a_n$ but by a rule that builds each term from the terms before it, together with enough starting terms to get going. Such a rule is called a recursive rule (or recurrence relation). For example, $$a_1=3,\qquad a_n=2a_{n-1}+1\ \ (n\ge2)$$ says: start with $3$; to get any later term, double the previous term and add $1$. So $a_2=7$, $a_3=15$, $a_4=31$, โฆ . A savings account works this way โ next month's balance is computed from this month's โ and so does a computer program that updates a value step by step.
The most famous recursive sequence needs two starting terms, because each term is the sum of the two before it.
The Fibonacci sequence $F_1,F_2,F_3,\dots$ is defined by $$F_1=1,\qquad F_2=1,\qquad F_n=F_{n-1}+F_{n-2}\ \ (n\ge3).$$ Its terms, the Fibonacci numbers, are $1,\ 1,\ 2,\ 3,\ 5,\ 8,\ 13,\ 21,\ 34,\ 55,\ 89,\ 144,\dots$ .
These numbers were found in India long before they reached Europe. A line of Sanskrit or Prakrit verse is built from short syllables (one beat) and long syllables (two beats). How many rhythms fill $n$ beats? A rhythm of $n$ beats ends either in a short syllable โ and before it is any rhythm of $n-1$ beats โ or in a long one โ and before it is any rhythm of $n-2$ beats. So the counts obey exactly the Fibonacci rule: $1$ rhythm of one beat, $2$ of two beats, then $3,5,8,13,\dots$ . The same argument counts the ways of climbing a staircase one or two steps at a time. Fibonacci numbers also appear in the spirals of sunflower heads, pineapples and pine cones.
(b) Write the first six terms of $a_1=1$, $a_2=3$, $a_n=a_{n-1}-a_{n-2}$ for $n\ge3$, and describe what happens.
Series and the Sigma Notation
If $a_1,a_2,a_3,\dots$ is a sequence, the expression $$a_1+a_2+a_3+\cdots$$ is called the series associated with it. A series is finite or infinite according as the sequence is finite or infinite. The sum of the first $n$ terms is the $n$th partial sum $$S_n=a_1+a_2+\cdots+a_n.$$ In sigma notation ($\Sigma$ is the Greek capital S, for "sum") this is written $$S_n=\sum_{k=1}^{n}a_k,$$ read "the sum of $a_k$ for $k$ from $1$ to $n$". The letter $k$ is the index; it takes every whole-number value from the lower limit to the upper limit, and the terms are added.
For instance $\displaystyle\sum_{k=1}^{4}k^2=1^2+2^2+3^2+4^2=30$ and $\displaystyle\sum_{k=3}^{6}(2k+1)=7+9+11+13=40$. The index is a "dummy" letter: $\displaystyle\sum_{k=1}^{4}k^2$ and $\displaystyle\sum_{j=1}^{4}j^2$ are the same sum. A sum from $k=m$ to $k=n$ has $n-m+1$ terms (not $n-m$ โ count $3,4,5,6$: four terms). The word series refers to the sum written out; its value is the number you get by adding.
For any numbers $a_1,\dots,a_n$ and $b_1,\dots,b_n$ and any constant $c$: $$\sum_{k=1}^{n}(a_k+b_k)=\sum_{k=1}^{n}a_k+\sum_{k=1}^{n}b_k,$$ $$\sum_{k=1}^{n}c\,a_k=c\sum_{k=1}^{n}a_k,\qquad \sum_{k=1}^{n}c=nc.$$
Proof. Write the left-hand sides out in full. In $(a_1+b_1)+(a_2+b_2)+\dots+(a_n+b_n)$ the $2n$ numbers can be added in any order, so collect the $a$'s and then the $b$'s. In $ca_1+ca_2+\dots+ca_n$ take out the common factor $c$. Finally $\sum_{k=1}^{n}c$ means $c+c+\dots+c$ with one $c$ for each of the $n$ values of $k$, which is $nc$. $\blacksquare$
If $S_n=a_1+a_2+\cdots+a_n$ for every $n\ge1$, then $$a_1=S_1,\qquad a_n=S_n-S_{n-1}\ \ (n\ge2).$$
Proof. $S_1$ is the sum of the first term alone. For $n\ge2$, $S_n=(a_1+\cdots+a_{n-1})+a_n=S_{n-1}+a_n$; subtract $S_{n-1}$. $\blacksquare$ The formula $S_n-S_{n-1}$ is valid only from $n=2$ on, so always check $a_1=S_1$ separately.
Recall: the arithmetic progression. In Class 10 you studied the sequence in which each term is the previous one plus a fixed number $d$, the common difference: $a,\ a+d,\ a+2d,\dots$ โ an arithmetic progression (AP). Its $n$th term and the sum of its first $n$ terms are $$a_n=a+(n-1)d,\qquad S_n=\tfrac{n}{2}\,[2a+(n-1)d]=\tfrac n2\,(a+l),$$ where $l=a_n$ is the last term. The sum is found by writing the series forwards and backwards and adding: every pair adds to $a+l$. The case $a=d=1$ gives the sum of the first $n$ natural numbers, $\displaystyle\sum_{k=1}^{n}k=\dfrac{n(n+1)}{2}$, which we shall use with Theorem 8.1. Recursively, an AP is $a_1=a$, $a_n=a_{n-1}+d$; in Section 8.2 the "$+d$" becomes "$\times r$".
(b) Write in sigma notation: (i) $2+5+8+11+\cdots+29$ (ii) $\dfrac12-\dfrac14+\dfrac18-\cdots-\dfrac1{64}$.
(b) The sum of the first $n$ terms of a sequence is $S_n=3n^2+2n$ for every $n$. Find $a_n$ and show that the sequence is an AP.
(c) Repeat (b) for $S_n=n^2+n+2$.
1. Miscounting the terms. $\displaystyle\sum_{k=4}^{10}$ has $10-4+1=7$ terms, not $6$. In particular $\displaystyle\sum_{k=4}^{10}5=35$.
2. Forgetting that a constant is added $n$ times. $\displaystyle\sum_{k=1}^{n}(2k+3)=n(n+1)+3n$, not $n(n+1)+3$.
3. Splitting products. $\displaystyle\sum a_kb_k\ne\Big(\sum a_k\Big)\Big(\sum b_k\Big)$ and $\displaystyle\sum a_k^2\ne\Big(\sum a_k\Big)^2$: for $k=1,2$, $1^2+2^2=5$ but $(1+2)^2=9$. Only sums and constant multiples split (Theorem 8.1).
4. Using $a_n=S_n-S_{n-1}$ at $n=1$. $S_0$ is not part of the data; find $a_1=S_1$ directly, as in Example 6(c).
5. Joining the dots. The graph of a sequence is a set of isolated points $(n,a_n)$; there are no terms between them.
The sums $\sum k$, $\sum k^2$, $\sum k^3$ and the principle of mathematical induction are taught and tested in school, but not in the year-end board exam. Of the entrance exams, only JEE Advanced lists the sums of squares and cubes of the first $n$ natural numbers (JEE Main lists AP and GP, so $\sum k$ appears there as an AP sum); induction is in neither JEE Main nor JEE Advanced.
Three special sums. $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2},\qquad \sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6},$$ $$\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2.$$ Why the middle one holds. Expand $(k+1)^3-k^3=3k^2+3k+1$ and add for $k=1,\dots,n$. On the left everything cancels except $(n+1)^3-1$. So $(n+1)^3-1=3\sum k^2+3\cdot\tfrac{n(n+1)}{2}+n$; solving for $\sum k^2$ and simplifying gives $\tfrac{n(n+1)(2n+1)}{6}$. The same trick with $(k+1)^4-k^4$ gives $\sum k^3$.
Worked example. Find $1\cdot2+2\cdot3+3\cdot4+\cdots+n(n+1)$.
Solution. $\displaystyle\sum_{k=1}^{n}k(k+1)=\sum k^2+\sum k=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(2n+4)}{6}=\frac{n(n+1)(n+2)}{3}.$ Check $n=3$: $2+6+12=20=\tfrac{3\cdot4\cdot5}{3}$ โ.
Mathematical induction. Checking a formula for $n=1,2,3,\dots,100$ never proves it for every $n$. Induction does, in two steps. Let $P(n)$ be a statement about the natural number $n$.
- Base step. Show that $P(1)$ is true.
- Inductive step. Show that for every natural number $k$, if $P(k)$ is true (the inductive hypothesis) then $P(k+1)$ is true.
Then $P(n)$ is true for all $n\in\mathbb{N}$: $P(1)$ holds, so $P(2)$ holds, so $P(3)$ holds, and so on โ like a row of dominoes, where the first one is pushed and each one knocks over the next. Both steps are needed. $P(n)$: "$n^2+n+41$ is prime" is true for $n=1,2,\dots,39$, but $40^2+40+41=1681=41^2$, so no inductive step can exist. And $P(n)$: "$2n+1$ is even" passes the inductive step (if $2k+1$ is even, so is $2k+3=(2k+1)+2$) yet is never true, because the base step fails.
Worked example (induction). Prove that $1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$ for all $n\in\mathbb{N}$.
Solution. Base: for $n=1$ the left side is $1$ and the right side is $\tfrac{1\cdot2\cdot3}{6}=1$ โ. Inductive step: assume $1^2+\cdots+k^2=\tfrac{k(k+1)(2k+1)}{6}$. Then $$1^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$ $$=\frac{(k+1)\big[k(2k+1)+6(k+1)\big]}{6}$$ $$=\frac{(k+1)(2k^2+7k+6)}{6}$$ $$=\frac{(k+1)(k+2)(2k+3)}{6},$$ which is the formula with $n=k+1$ (since $2(k+1)+1=2k+3$). By induction the formula holds for every $n$.
Worked example (divisibility). Prove that $3^{2n}-1$ is divisible by $8$ for all $n\in\mathbb{N}$.
Solution. Base: $3^2-1=8$ โ. Inductive step: assume $9^k-1=8m$ for some integer $m$ (note $3^{2k}=9^k$). Then $9^{k+1}-1=9\cdot9^k-1=9(8m+1)-1=72m+8=8(9m+1)$, a multiple of $8$. Hence $8$ divides $3^{2n}-1$ for every $n$. Check $n=2$: $81-1=80=8\times10$ โ.
Exercises
- 1 Write the first five terms of $a_n=2n+5$.
- 2 Write the first five terms of $a_n=\dfrac{n}{n+1}$.
- 3 Write the first five terms of $a_n=(-3)^n$.
- 4 Write the first three terms of $a_n=\dfrac{2n-3}{6}$.
- 5 Find $a_{10}$ for $a_n=n^2-n+1$.
- 6 Find $a_7$ and $a_8$ for $a_n=(-1)^{n-1}n^3$.
- 7 Which term of the sequence $a_n=n^2-3n$ is $40$?
- 8 Is $150$ a term of the sequence $a_n=n(n+1)$? Give a reason.
- 1 Find a formula for the $n$th term of $4,7,10,13,\dots$ .
- 2 Find a formula for the $n$th term of $1,\tfrac13,\tfrac19,\tfrac1{27},\dots$ .
- 3 Find a formula for the $n$th term of $\tfrac12,-\tfrac23,\tfrac34,-\tfrac45,\dots$ .
- 4 Find a formula for the $n$th term of $2,6,12,20,30,\dots$ .
- 5 Write the first five terms of $a_1=2$, $a_n=a_{n-1}+3$ ($n\ge2$). What kind of sequence is it?
- 6 Write the first five terms of $a_1=-1$, $a_n=\dfrac{a_{n-1}}{n}$ ($n\ge2$).
- 7 Write the first six terms of $a_1=1$, $a_2=2$, $a_n=a_{n-1}\,a_{n-2}$ ($n\ge3$).
- 8 For the Fibonacci sequence, find $F_{10}$ and verify that $F_1+F_2+\cdots+F_8=F_{10}-1$.
- 1 Expand and evaluate $\displaystyle\sum_{k=1}^{6}(3k-2)$.
- 2 Evaluate $\displaystyle\sum_{k=1}^{5}2^k$.
- 3 Evaluate $\displaystyle\sum_{k=3}^{7}(-1)^k\,k$.
- 4 Write $1+4+7+\cdots+40$ in sigma notation.
- 5 Write $\tfrac13+\tfrac19+\tfrac1{27}+\cdots+\tfrac1{729}$ in sigma notation.
- 6 Write $2\cdot3+3\cdot4+4\cdot5+\cdots+20\cdot21$ in sigma notation.
- 7 Use Theorem 8.1 to evaluate $\displaystyle\sum_{k=1}^{10}(2k+5)$.
- 8 If $\displaystyle\sum_{k=1}^{12}a_k=40$ and $\displaystyle\sum_{k=1}^{12}b_k=15$, find $\displaystyle\sum_{k=1}^{12}(3a_k-2b_k+1)$.
- 9 The sum of the first $n$ terms of a sequence is $S_n=n^2+4n$. Find $a_n$. Is the sequence an AP?
- 10 The sum of the first $n$ terms of a sequence is $S_n=2^n-1$. Find $a_n$.
- 11 The sum of the first $n$ terms of a sequence is $S_n=n^2+n+1$. Find $a_1$ and $a_n$ for $n\ge2$. Is the sequence an AP?
Type the term with k, + - * /, powers ^ and brackets, e.g. (-1)^k*k or 2^k.
- 1 An auditorium has $25$ rows. The first row has $20$ seats and each row has $2$ more seats than the row in front. How many seats are in the last row, and how many in total?
- 2 A ball is dropped from a height of $16\ \text{m}$. After each bounce it rises to $\tfrac34$ of the height it fell from. Write a recursive rule for the height $h_n$ reached after the $n$th bounce, and find $h_1,\dots,h_4$.
- 3 A lake holds $5000$ fish. Each year the number grows by $10\%$, and then $300$ fish are caught. Write a recursive rule for the number $P_n$ after $n$ years ($P_0=5000$) and find $P_3$.
- 4 You climb a staircase taking either one or two steps at a time. Let $c_n$ be the number of ways to climb $n$ steps. Explain why $c_n=c_{n-1}+c_{n-2}$ for $n\ge3$, and find $c_8$.
- 5 Find the sum of all natural numbers between $100$ and $200$ that are divisible by $7$.
- 6 Rina saves โน50 in the first week, and each week she saves โน10 more than the week before. In which week does her total savings first reach โน2000?
- 1 Evaluate $\displaystyle\sum_{k=1}^{10}k^2$ and $\displaystyle\sum_{k=1}^{10}k^3$.
- 2 Prove by induction that $1+3+5+\cdots+(2n-1)=n^2$ for all $n\in\mathbb{N}$.
- 3 Prove by induction that $1^3+2^3+\cdots+n^3=\left[\dfrac{n(n+1)}{2}\right]^2$ for all $n\in\mathbb{N}$.
- 4 Prove by induction that $5^n-1$ is divisible by $4$ for all $n\in\mathbb{N}$.
- 5 Prove by induction that $2^n\gt n$ for all $n\in\mathbb{N}$.
- 6 Show that $1\cdot2\cdot3+2\cdot3\cdot4+\cdots+n(n+1)(n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$.
How this is examined: Board papers ask for the first few terms of a sequence given by a formula or a recursive rule (1โ2 marks) and use $a_n=S_n-S_{n-1}$ (2โ3 marks); case studies build a recursion from a story. JEE uses recursions whose pattern you must spot, and identities between Fibonacci-type numbers.
E1. Board ยท 1 mark The $5$th term of the sequence $a_n=\dfrac{(-1)^{n-1}\,2^n}{n}$ is (A) $-\dfrac{32}{5}$ (B) $\dfrac{32}{5}$ (C) $\dfrac{16}{5}$ (D) $-\dfrac{16}{5}$
E2. Board ยท 2 marks Write the first five terms of the sequence $a_1=1$, $a_n=3a_{n-1}+1$ ($n\ge2$), and the corresponding series.
E3. Board ยท 3 marks The sum of the first $n$ terms of a sequence is $S_n=4n^2-n$. Find its $n$th term, show that it is an AP, and find $a_{10}$ and $a_{11}+a_{12}+\cdots+a_{20}$.
E4. Board ยท Case study A garden path $2$ units wide and $n$ units long is to be paved with identical rectangular tiles of size $1\times2$, laid lengthwise or crosswise, with no gaps or overlaps. Let $t_n$ be the number of ways; then $t_1=1$, $t_2=2$. (i) Find $t_3$ and $t_4$ by drawing. (1 mark) (ii) By looking at the tiles at the end of the path, explain why $t_n=t_{n-1}+t_{n-2}$ for $n\ge3$. (1 mark) (iii) Find $t_{10}$. (2 marks)
E5. JEE Main A sequence satisfies $a_1=1$ and $a_{n+1}=\dfrac{a_n}{1+a_n}$ for $n\ge1$. Then $a_{10}$ equals (A) $\dfrac1{10}$ (B) $\dfrac19$ (C) $\dfrac1{11}$ (D) $\dfrac1{2^9}$
E6. JEE Advanced One or more options correct. For the Fibonacci numbers $F_1=F_2=1$, $F_n=F_{n-1}+F_{n-2}$: (A) $F_1+F_2+\cdots+F_n=F_{n+2}-1$ for all $n\ge1$ (B) $F_{n+1}F_{n-1}-F_n^2=1$ for all $n\ge2$ (C) $F_1+F_3+F_5+\cdots+F_{2n-1}=F_{2n}$ for all $n\ge1$ (D) $F_n$ is even exactly when $n$ is a multiple of $3$
Research question. Do the spirals on pine cones, pineapples and flower heads really come in Fibonacci numbers โ and how often?
Why it matters. Plants grow new seeds and leaves one at a time, each turned by about the same angle from the last. Botanists and mathematicians have shown that this simple recursive growth rule packs seeds most efficiently when the angle is near $137.5^\circ$, and the visible spirals then come in neighbouring Fibonacci numbers. Checking a mathematical claim against real data โ and counting the exceptions honestly โ is what scientists do.
What to do.
- Collect $5$โ$10$ specimens: pine cones, a pineapple, a sunflower or marigold head, a cauliflower or a romanesco (photographs from the market are fine).
- On each, mark one spiral with a thread or a felt pen and count all the spirals running clockwise; then count those running anticlockwise.
- Record the pairs in a table and mark which numbers are Fibonacci numbers ($1,2,3,5,8,13,21,34,55,89,\dots$).
- For each specimen compute the ratio larger รท smaller and compare it with $1.618$.
Tools: Specimens or close-up phone photos, thread or a marker, a spreadsheet
Hand in: The table of spiral counts, a photo of one marked specimen, and a paragraph: what fraction of your specimens gave neighbouring Fibonacci numbers, and what could explain the others?
Going further. Put dots at distances $\sqrt{n}$ from a centre, each turned $137.5^\circ$ from the previous one, for $n=1,\dots,300$ (a spreadsheet chart or a graphing app will do). Count the spirals you see. Now try $137^\circ$ and $140^\circ$ โ what changes?
๐งช Self-check quiz
Q1. If $a_n=n^2-2n$, then $a_4$ is
Q2. The $7$th Fibonacci number $F_7$ (with $F_1=F_2=1$) is
Q3. $\displaystyle\sum_{k=1}^{4}(2k+1)$ equals
Q4. The sum of the first $n$ terms of a sequence is $S_n=n^2+2n$. Its $5$th term is
Q5. $\displaystyle\sum_{k=1}^{n}5$ is equal to