MathematicsClass 9CBSE

One Rule Gives You the Hundredth Term; the Other Makes You Walk There

Learn to spot the pattern in a sequence and continue it, write the explicit rule for the nth term, write the recursive rule including the Fibonacci sequence, and convert between the two descriptions.

What are the two different ways to describe a sequence?

Look at

2

5

8

11

14

. There are two honest ways to say what is going on.

The recursive way: start at

2

and keep adding

3

a_{1} = 2, a_{n} = a_{n - 1} + 3

The explicit way: the

n

th term is three times

n

, less one.

a_{n} = 3 n - 1

Both describe the same list. But ask for the hundredth term and they behave completely differently. The explicit rule gives it at once:

a_{100} = 3 (100) - 1 = 299

The recursive rule gives it too — after you have worked out the other ninety-nine terms first.

So the explicit rule is the more useful one whenever it exists. Its drawback is that some sequences have no simple explicit rule at all, and the Fibonacci sequence in the third section is the standard example: it is easy to describe recursively and hard to describe explicitly.

This page covers the first part of the CBSE Class 9 Mathematics chapter on sequences — continuing a pattern, writing both kinds of rule, and moving between them.

How do you spot the pattern and write the next three terms?

Write the differences between consecutive terms first. If they are constant, the rule is add the same amount. If they are not, look at the ratios, then at the differences of the differences.

Worked example 1 — constant differences.

2

5

8

11

. Differences:

3

3

3

. So keep adding

3

14, 17, 20

Worked example 2 — constant ratio.

3

6

12

24

. Differences

3

6

12

are not constant, but the ratios are

2

2

2

. So keep doubling:

48, 96, 192

Worked example 3 — square numbers.

1

4

9

16

. Differences

3

5

7

are not constant, but they grow by

2

each time, so the next differences are

9

11

13

25, 36, 49

These are

5^{2}

6^{2}

7^{2}

— the square numbers.

Worked example 4 — triangular numbers.

1

3

6

10

. Differences

2

3

4

, increasing by one, so the next differences are

5

6

7

15, 21, 28

These count dots stacked in a triangle: one row of

1

, then

1 + 2

, then

1 + 2 + 3

, and so on.

Worked example 5 — each term from the two before it.

1

1

2

3

5

. Differences are

0

1

1

2

, which is the sequence itself shifted — a signal to try adding pairs. Indeed

1 + 1 = 2

1 + 2 = 3

2 + 3 = 5

, so

8, 13, 21

This is the Fibonacci sequence.

Worked example 6 — an alternating sign.

2

- 4

8

- 16

. The ratio is

- 2

each time:

32, - 64, 128

A finite list never forces a single continuation. After

1

2

4

the answer

8

is natural under keep doubling, but

7

is equally defensible under add one more each time. So a complete answer states the rule as well as the terms — the rule is what makes the continuation more than a guess.

Where the structure supplies the certainty. For the triangular numbers the rule is not fitted to the numbers at all; it is a count of dots, and each new row genuinely adds one more dot than the row before. That is the difference between a pattern you have justified and one you have merely matched, and the next sections turn justified patterns into formulas.

Formula

How do you write the explicit rule for the nth term?

**For a sequence with constant differences, the difference is the coefficient of

n

**, and the constant is whatever makes the first term come out right:

a_{n} = d n + (a_{1} - d)

For sequences that are not of that kind, recognise the standard families below.

**Worked example 1 — constant difference

3

.** For

2

5

8

11

d = 3

, and

a_{1} - d = 2 - 3 = - 1

, so

a_{n} = 3 n - 1

Check at

n = 1

2

. At

n = 4

11

. Then

a_{20} = 60 - 1 = 59

and

a_{100} = 299

.

**Worked example 2 — constant difference

4

.** For

5

9

13

17

d = 4

and

a_{1} - d = 1

, so

a_{n} = 4 n + 1

Check at

n = 1

5

. Then

a_{20} = 81

.

Worked example 3 — square numbers. For

1

4

9

16

a_{n} = n^{2} \Rightarrow a_{20} = 400

Worked example 4 — triangular numbers. The

n

th triangular number is the sum

1 + 2 + \dots + n

, and the closed form is

T_{n} = \frac{n ( n + 1 )}{2}

Check at

n = 4

\frac{4 \times 5}{2} = 10

. Correct. Then

T_{10} = \frac{10 \times 11}{2} = 55

and

T_{20} = \frac{20 \times 21}{2} = 210

.

Worked example 5 — a doubling sequence. For

3

6

12

24

the ratio is

2

, so

a_{n} = 3 \times 2^{n - 1}

Check at

n = 1

3 \times 2^{0} = 3

. At

n = 4

3 \times 8 = 24

. Then

a_{10} = 3 \times 512 = 1536

.

Worked example 6 — working backwards from a term. Which term of

2

5

8

11

equals

59

? Solve

3 n - 1 = 59

3 n = 60 \Rightarrow n = 20

59

is the twentieth term. And is

100

a term?

3 n - 1 = 100

gives

3 n = 101

, which is not divisible by

3

— so no.

Always check the rule at a term other than the first. The constant was chosen to make

a_{1}

correct, so the first term proves nothing. The second or third term is the real test, and it takes one substitution.

Not every sequence has a workable explicit rule. The Fibonacci sequence does have one, but it involves irrational numbers and is well beyond this stage — so it is described recursively instead, which is the next section.

How do you write a recursive rule and use it for the Fibonacci sequence?

Give the starting term or terms, then a rule that builds each new term from the ones before it. A recursive description always needs both halves.

Worked example 1 — adding a constant. For

2

5

8

11

a_{1} = 2, a_{n} = a_{n - 1} + 3 (n > 1)

Generating:

a_{2} = 5

a_{3} = 8

a_{4} = 11

a_{5} = 14

. Correct.

Worked example 2 — multiplying by a constant. For

3

6

12

24

a_{1} = 3, a_{n} = 2 a_{n - 1}

Generating:

3

6

12

24

48

.

Worked example 3 — triangular numbers. Each one adds the next whole number:

T_{1} = 1, T_{n} = T_{n - 1} + n

Generating:

T_{2} = 1 + 2 = 3

T_{3} = 3 + 3 = 6

T_{4} = 6 + 4 = 10

T_{5} = 15

. Correct.

Worked example 4 — square numbers. Each square adds the next odd number:

S_{1} = 1, S_{n} = S_{n - 1} + (2 n - 1)

Generating:

S_{2} = 1 + 3 = 4

S_{3} = 4 + 5 = 9

S_{4} = 9 + 7 = 16

. Correct — and this is why

1 + 3 + 5 + 7 = 16

, a fact the next part of this chapter uses.

Worked example 5 — the Fibonacci sequence. Here each term needs the two before it, so two starting values are required:

F_{1} = 1, F_{2} = 1, F_{n} = F_{n - 1} + F_{n - 2}

Generating step by step:

-

F_{3} = 1 + 1 = 2

F_{4} = 1 + 2 = 3

F_{5} = 2 + 3 = 5

F_{6} = 3 + 5 = 8

F_{7} = 5 + 8 = 13

F_{8} = 8 + 13 = 21

F_{9} = 13 + 21 = 34

F_{10} = 21 + 34 = 55

The number of starting values must match how far back the rule reaches. A rule using only

a_{n - 1}

needs one starting value; Fibonacci's rule reaches back two terms, so it needs two. **Giving only

F_{1} = 1

would leave the sequence undetermined**, since

F_{2}

could be anything and every later term would change with it.

Worked example 6 — the same rule, different start. Keep

F_{n} = F_{n - 1} + F_{n - 2}

but begin

2

1

2, 1, 3, 4, 7, 11, 18, 29

A completely different list from the identical rule. So the starting values are part of the description, not preamble to it.

Recursion is how a growing pattern actually behaves. A matchstick row genuinely gains three sticks per square, and a stack of dots genuinely gains one more row — the recursive rule states the mechanism. The explicit rule is the shortcut derived from it, which is why converting between them is worth being able to do in both directions.

How do you convert between the explicit and recursive descriptions?

Explicit to recursive: find the difference between consecutive terms. Recursive to explicit: identify the constant difference and fit the constant.

Worked example 1 — explicit to recursive. Given

a_{n} = 3 n - 1

, compute the difference:

a_{n} - a_{n - 1} = (3 n - 1) - (3 (n - 1) - 1) = 3 n - 1 - 3 n + 3 + 1 = 3

So the recursive form is

a_{n} = a_{n - 1} + 3

, with

a_{1} = 3 (1) - 1 = 2

.

Worked example 2 — recursive to explicit. Given

a_{1} = 5

and

a_{n} = a_{n - 1} + 4

, the difference is

4

and the constant is

5 - 4 = 1

a_{n} = 4 n + 1

Check at

n = 1

5

. At

n = 3

13

, matching

5

9

13

.

Worked example 3 — explicit to recursive for the squares. Given

S_{n} = n^{2}

S_{n} - S_{n - 1} = n^{2} - (n - 1)^{2} = n^{2} - n^{2} + 2 n - 1 = 2 n - 1

S_{n} = S_{n - 1} + (2 n - 1)

with

S_{1} = 1

— the odd numbers, as the previous section found by inspection.

Worked example 4 — explicit to recursive for the triangular numbers. Given

T_{n} = \frac{n ( n + 1 )}{2}

T_{n} - T_{n - 1} = \frac{n ( n + 1 )}{2} - \frac{( n - 1 ) n}{2} = \frac{n [ ( n + 1 ) - ( n - 1 ) ]}{2} = \frac{2 n}{2} = n

T_{n} = T_{n - 1} + n

with

T_{1} = 1

. Both descriptions confirmed against each other.

Worked example 5 — recognising a family from its rule. A sequence has

a_{n} = a_{n - 1} + (2 n - 1)

with

a_{1} = 1

. The increments are the odd numbers, so the terms are the partial sums of the odds — the square numbers, and the explicit rule is

a_{n} = n^{2}

.

Another has

a_{n} = a_{n - 1} + n

with

a_{1} = 1

. The increments are the whole numbers, so these are the triangular numbers,

T_{n} = \frac{n ( n + 1 )}{2}

.

Worked example 6 — a relation between the two families. Add two consecutive triangular numbers:

T_{4} + T_{3} = 10 + 6 = 16 = 4^{2}

T_{5} + T_{4} = 15 + 10 = 25 = 5^{2}

T_{n} + T_{n - 1} = n^{2}

. Check algebraically:

\frac{n ( n + 1 )}{2} + \frac{( n - 1 ) n}{2} = \frac{n [ ( n + 1 ) + ( n - 1 ) ]}{2} = \frac{2 n ^{2}}{2} = n^{2}

Two triangles of dots, one turned upside down, fit together into a square — which is the picture that identity draws.

Conversion is only easy in one direction for some sequences. Going explicit to recursive is always a subtraction and always works. Going recursive to explicit requires recognising the pattern of increments, and for Fibonacci no elementary explicit rule exists — the increments are the sequence itself, so there is nothing simpler to sum. That asymmetry is the reason both descriptions are taught, rather than one being treated as the proper form.

Exam tip

Exam tip: write the differences first, then check at a second term

Write the differences under the sequence before anything else. Constant differences mean a linear rule; growing differences mean squares or triangles; constant ratios mean doubling or halving.

If the differences are not constant, take their differences too. For

1, 4, 9, 16

the second differences are all

2

— the signature of

n^{2}

.

State the rule, not just the next terms. *Add

3

each time* or

a_{n} = 3 n - 1

earns the marks; three numbers alone do not.

Check the explicit rule at a term other than the first — the constant was fitted to

a_{1}

, so

a_{2}

a_{3}

is the real test.

A recursive description needs its starting value. One starting value for a rule using

a_{n - 1}

; two for Fibonacci, which reaches back twice.

Show the generation step by step for a recursive question:

F_{3} = 2

F_{4} = 3

F_{5} = 5

.

**For "which term equals

k

", solve the rule as an equation**:

3 n - 1 = 59

gives

n = 20

. If

n

comes out fractional, the value is not a term — say so.

Learn the two standard forms:

T_{n} = \frac{n ( n + 1 )}{2}

for triangular numbers,

S_{n} = n^{2}

for squares.

**To convert explicit to recursive, subtract

a_{n - 1}

from

a_{n}

and simplify — it is always a short algebra step.

And remember

T_{n} + T_{n - 1} = n^{2}

** — it turns several triangular-number questions into one line.

Did you know

Why eight times a triangular number is always one short of a square

Take any triangular number, multiply it by eight, and add one. The result is always a perfect square.

-

T_{1} = 1

8 (1) + 1 = 9 = 3^{2}

T_{2} = 3

8 (3) + 1 = 25 = 5^{2}

T_{3} = 6

8 (6) + 1 = 49 = 7^{2}

T_{4} = 10

8 (10) + 1 = 81 = 9^{2}

T_{10} = 55

8 (55) + 1 = 441 = 2 1^{2}

The squares produced are

3^{2}, 5^{2}, 7^{2}, 9^{2}

— the odd squares, in order, with

T_{10}

giving

2 1^{2}

.

The algebra is short enough to check. With

T_{n} = \frac{n ( n + 1 )}{2}

8 T_{n} + 1 = 8 \times \frac{n ( n + 1 )}{2} + 1 = 4 n (n + 1) + 1 = 4 n^{2} + 4 n + 1 = (2 n + 1)^{2}

And

2 n + 1

is odd for every whole

n

, which is exactly the pattern observed. For

n = 10

it gives

21

, matching

441

.

There is also a picture. Eight copies of a triangle of dots can be arranged around a single central dot to fill a square — the one extra dot is the

+ 1

, and it sits in the middle where the eight triangles cannot reach.

What makes this worth noticing is that the identity is an algebraic rearrangement of a formula you already have. The expression

4 n^{2} + 4 n + 1

is a perfect square because of the identity

(a + b)^{2} = a^{2} + 2 ab + b^{2}

from the algebra chapter, with

a = 2 n

and

b = 1

.

So a fact about stacked dots turns out to be the square of a binomial, and the connection runs through

T_{n} = \frac{n ( n + 1 )}{2}

— a formula the next part of this chapter derives properly rather than quoting.

Exam relevance

How do sequences feed into JEE Main?

Because the explicit and recursive descriptions become the two standard ways of specifying a series, and recursion is the engine of mathematical induction.

This is the foundation for Class 11 Mathematics Sequences and Series and Principle of Mathematical Induction, both examined in JEE Main. The explicit rule

a_{n} = d n + (a_{1} - d)

found here is the

n

th term of an arithmetic progression, written there as

a_{n} = a + (n - 1) d

— the same expression rearranged. The doubling rule

a_{n} = 3 \times 2^{n - 1}

is the

n

th term of a geometric progression. Both standard forms are already on this page under different names, which the next two parts of this chapter make explicit.

Recursion is what induction proves things about. A recursive rule says how each term follows from the last, and induction is the method for proving a closed formula matches it — assume it for

n - 1

, derive it for

n

. The conversion worked through here, showing

T_{n} - T_{n - 1} = n

, is precisely the inductive step for the formula

T_{n} = \frac{n ( n + 1 )}{2}

, and Class 11 questions ask for exactly that argument written formally.

Triangular and square numbers reappear as summation formulas. Class 11 gives

k = 1 \sum n k = \frac{n ( n + 1 )}{2} k = 1 \sum n k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

and the first of those is the triangular-number formula. Questions requiring a sum of a series in

n

use these as building blocks, and the identity

T_{n} + T_{n - 1} = n^{2}

is the kind of manipulation that shortens them.

The Fibonacci sequence appears in JEE Advanced and olympiad problems as a recurrence to be analysed, and it is the standard example of why a recursive definition is sometimes the only practical one. Its explicit formula involves irrational numbers and is met much later, which is the honest reason it is left recursive here.

For NTSE and olympiad papers, pattern-continuation and next-term questions appear directly, and the second-differences technique on this page is the reliable method for them.

What the questions look like. For board work, expect write the next three terms with the rule stated, **find the

n

th term and a specified term such as the twentieth, generate terms from a recursive rule, convert between the two forms, and identify triangular or square numbers. For JEE Main, sequences appear as AP and GP problems, as summations, and as recurrences.

How board and competitive emphasis differ. A board paper rewards the stated rule** and the step-by-step generation. A competitive paper assumes the

n

th term formula and tests whether you can sum a series or recognise which progression is hiding in a word problem.

The single trap that costs the most marks. Giving a recursive rule without its starting values, or giving only one where two are needed.

a_{n} = a_{n - 1} + a_{n - 2}

with

F_{1} = 1

alone describes infinitely many sequences —

1, 1, 2, 3, 5

and

1, 4, 5, 9, 14

both obey it. The starting values are part of the answer, and an otherwise perfect rule loses marks without them.

Key takeaways

Sequences, explicit rules and recursion: quick revision

- Write the differences first. Constant differences mean a linear rule; constant ratios mean doubling; growing differences mean squares or triangles.
-

2, 5, 8, 11 \to 14, 17, 20

(add

3

3, 6, 12, 24 \to 48, 96, 192

(double).

2, - 4, 8, - 16 \to 32, - 64, 128

(ratio

- 2

).
-

1, 4, 9, 16 \to 25, 36, 49

— square numbers, second differences all

2

.
-

1, 3, 6, 10 \to 15, 21, 28

— triangular numbers, differences

2, 3, 4, 5

.
-

1, 1, 2, 3, 5 \to 8, 13, 21

— Fibonacci, each term the sum of the two before.
- State the rule, not just the terms — a finite list never forces one continuation.
- Explicit rule for constant difference

d

a_{n} = d n + (a_{1} - d)

.
-

2, 5, 8, 11

gives

a_{n} = 3 n - 1

, so

a_{20} = 59

and

a_{100} = 299

.
-

5, 9, 13, 17

gives

a_{n} = 4 n + 1

, so

a_{20} = 81

.
- Squares:

S_{n} = n^{2}

, so

S_{20} = 400

. Triangles:

T_{n} = \frac{n ( n + 1 )}{2}

, so

T_{10} = 55

and

T_{20} = 210

.
- Doubling:

3, 6, 12, 24

gives

a_{n} = 3 \times 2^{n - 1}

, so

a_{10} = 1536

.
- **"Which term equals

k

"**: solve

3 n - 1 = 59

to get

n = 20

3 n - 1 = 100

gives no whole

n

, so

100

is not a term.
- Check the explicit rule at a second term, since the constant was fitted to the first.
- Recursive rule needs a starting value and a step:

a_{1} = 2

a_{n} = a_{n - 1} + 3

.
-

T_{1} = 1

T_{n} = T_{n - 1} + n

generates the triangles;

S_{1} = 1

S_{n} = S_{n - 1} + (2 n - 1)

generates the squares by adding odd numbers.
- Fibonacci needs TWO starting values:

F_{1} = F_{2} = 1

with

F_{n} = F_{n - 1} + F_{n - 2}

, giving

1, 1, 2, 3, 5, 8, 13, 21, 34, 55

.
- The same rule from

2, 1

gives

2, 1, 3, 4, 7, 11, 18, 29

— the starting values are part of the description.
- Explicit to recursive: subtract.

(3 n - 1) - (3 (n - 1) - 1) = 3

;

n^{2} - (n - 1)^{2} = 2 n - 1

;

T_{n} - T_{n - 1} = n

.
- **

T_{n} + T_{n - 1} = n^{2}

**:

10 + 6 = 16

and

15 + 10 = 25

, proved by

\frac{n [( n + 1 ) + ( n - 1 )]}{2} = n^{2}

.
- **

8 T_{n} + 1

is always an odd square**:

8 (10) + 1 = 81 = 9^{2}

, since

4 n^{2} + 4 n + 1 = (2 n + 1)^{2}

.
- Fibonacci has no elementary explicit rule at this stage, which is why both descriptions are needed.

Write out the Fibonacci sequence to the fifteenth term, then divide each term by the one before it and watch what the answers settle towards.

What are the two different ways to describe a sequence?

How do you spot the pattern and write the next three terms?

How do you write the explicit rule for the nth term?

How do you write a recursive rule and use it for the Fibonacci sequence?

How do you convert between the explicit and recursive descriptions?

Exam tip: write the differences first, then check at a second term

Why eight times a triangular number is always one short of a square

How do sequences feed into JEE Main?

Sequences, explicit rules and recursion: quick revision

Related study notes

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