Sequences and Series

Numbers in a Definite Order

How many ancestors do you have over 300 years? With a 30-year generation gap, the generations contribute $2, 4, 8, 16, \ldots, 1024$ people — numbers arranged in a definite order according to a rule. That is a sequence. The successive quotients of $10 \div 3$ — $3, 3.3, 3.33, 3.333, \ldots$ — form another.

Key Point: The numbers in a sequence are its terms, written $a_1, a_2, a_3, \ldots$ — the subscript records the position. The $n$-th term $a_n$ is the general term. A sequence with a fixed number of terms is finite (the 10 generations); otherwise it is infinite (the quotients never end).

Formally, a sequence is a function whose domain is $\mathbb{N}$ (or a subset $\{1, 2, \ldots, k\}$) — sometimes written $a(n)$ for $a_n$. Position in, term out.

[Board Tip] "Write the first three terms of $a_n = 2n + 5$" means substitute $n = 1, 2, 3$: the terms are $7, 9, 11$. Substitution IS the method — no cleverness needed.

Three Kinds of Rules — and Series

A sequence's rule can be:

• A formula: even naturals $a_n = 2n$; odd naturals $a_n = 2n - 1$.
• A recurrence — each term built from earlier ones. The Fibonacci sequence: $a_1 = a_2 = 1$ and $a_n = a_{n-1} + a_{n-2}$ for $n > 2$, giving $1, 1, 2, 3, 5, 8, 13, \ldots$
• A verbal description only: the primes $2, 3, 5, 7, \ldots$ — no formula for the $n$-th prime exists.

Key Point (Series): From a sequence $a_1, a_2, a_3, \ldots$, the expression $a_1 + a_2 + a_3 + \cdots$ is the associated series, compactly $\sum_{k=1}^{n} a_k$ in sigma notation. The word "series" names the indicated sum; the "sum of the series" is its value — $1 + 3 + 5 + 7$ is a four-term series whose sum is 16.

[JEE Tip] Recursive definitions must be unrolled in order — to get $a_5$ of $a_1 = 3$, $a_n = 3a_{n-1} + 2$, compute $a_2, a_3, a_4$ first ($11, 35, 107$), then $a_5 = 323$. There is no shortcut without solving the recurrence.

Solved Examples

Example 1: Terms by substitution

Write the first three terms of (i) $a_n = 2n + 5$ (ii) $a_n = \frac{n-3}{4}$.

Solution:

Step 1 — (i) Substitute $n = 1, 2, 3$. $a_1 = 2 + 5 = 7$, $a_2 = 4 + 5 = 9$, $a_3 = 6 + 5 = 11$.

Step 2 — (ii) Substitute again. $a_1 = \frac{1-3}{4} = -\frac{1}{2}$, $a_2 = \frac{2-3}{4} = -\frac{1}{4}$, $a_3 = \frac{3-3}{4} = 0$.

Takeaway: A formula rule needs nothing but substitution — position in, term out.

Example 2: A distant term

What is the 20th term of the sequence $a_n = (n-1)(2-n)(3+n)$?

Solution:

Step 1 — Substitute $n = 20$ into each factor. $n - 1 = 19$, $2 - n = -18$, $3 + n = 23$.

Step 2 — Multiply. $19 \times (-18) = -342$; $-342 \times 23 = -7866$.

Takeaway: The middle factor $(2-n)$ is negative for every $n > 2$ — expect negative distant terms.

Example 3: Recurrence and its series

The sequence is defined by $a_1 = 1$, $a_n = a_{n-1} + 2$ for $n \ge 2$. Find the first five terms and write the corresponding series.

Solution:

Step 1 — Unroll the recurrence in order. $a_2 = 1 + 2 = 3$, $a_3 = 5$, $a_4 = 7$, $a_5 = 9$.

Step 2 — Recognise the pattern. These are the odd numbers: $a_n = 2n - 1$.

Step 3 — Write the series. $1 + 3 + 5 + 7 + 9 + \cdots$

Takeaway: "Add 2 each time" from 1 generates exactly the odd numbers — recurrence and formula describe one sequence.

Example 4: Formula drills

Write the first five terms of (i) $a_n = n(n+2)$ (ii) $a_n = 2^n$.

Solution:

Step 1 — (i) Substitute $n = 1$ to $5$. $1 \cdot 3, \ 2 \cdot 4, \ 3 \cdot 5, \ 4 \cdot 6, \ 5 \cdot 7 = 3, 8, 15, 24, 35$.

Step 2 — (ii) Powers of 2. $2, 4, 8, 16, 32$.

Takeaway: Writing a couple of factored forms before multiplying keeps the arithmetic honest.

Example 5: Alternating signs

Write the first five terms of $a_n = (-1)^{n-1}\,5^{n+1}$.

Solution:

Step 1 — Read the two factors. $(-1)^{n-1}$ controls the sign (positive at $n = 1$); $5^{n+1}$ controls the size.

Step 2 — Substitute. $a_1 = +5^2 = 25$, $a_2 = -5^3 = -125$, $a_3 = +5^4 = 625$, $a_4 = -3125$, $a_5 = 15625$.

Takeaway: A $(-1)^{n-1}$ factor alternates signs starting positive; $(-1)^n$ would start negative.

Example 6: Indicated terms

Find (i) $a_{17}$ and $a_{24}$ for $a_n = 4n - 3$ (ii) $a_7$ for $a_n = \frac{n^2}{2^n}$ (iii) $a_9$ for $a_n = (-1)^{n-1}n^3$ (iv) $a_{20}$ for $a_n = \frac{n(n-2)}{n+3}$.

Solution:

Step 1 — (i). $a_{17} = 68 - 3 = 65$; $a_{24} = 96 - 3 = 93$.

Step 2 — (ii). $a_7 = \frac{49}{128}$.

Step 3 — (iii). $n = 9$ makes $(-1)^{8} = +1$: $a_9 = 729$.

Step 4 — (iv). $a_{20} = \frac{20 \times 18}{23} = \frac{360}{23}$.

Takeaway: Every part is one substitution — the only care point is evaluating the sign factor correctly.

Example 7: Recurrences unrolled

Write the first five terms of (i) $a_1 = 3$, $a_n = 3a_{n-1} + 2$ (ii) $a_1 = -1$, $a_n = \frac{a_{n-1}}{n}$.

Solution:

Step 1 — (i) Unroll in order. $a_2 = 3(3)+2 = 11$, $a_3 = 3(11)+2 = 35$, $a_4 = 3(35)+2 = 107$, $a_5 = 3(107)+2 = 323$. Series: $3 + 11 + 35 + 107 + 323 + \cdots$

Step 2 — (ii) Divide by the index each time. $a_2 = -\frac{1}{2}$, $a_3 = -\frac{1}{6}$, $a_4 = -\frac{1}{24}$, $a_5 = -\frac{1}{120}$.

Step 3 — Spot the pattern in (ii). The denominators are $1, 2, 6, 24, 120 = n!$: $a_n = -\frac{1}{n!}$.

Takeaway: Recurrences have no shortcut — each term needs the one before it, computed in order.

Example 8: Fibonacci ratios

For the Fibonacci sequence, find $\frac{a_{n+1}}{a_n}$ for $n = 1, \ldots, 5$.

Solution:

Step 1 — List the terms. $1, 1, 2, 3, 5, 8$.

Step 2 — Form the successive ratios. $\frac{1}{1} = 1$, $\frac{2}{1} = 2$, $\frac{3}{2} = 1.5$, $\frac{5}{3} \approx 1.67$, $\frac{8}{5} = 1.6$.

Takeaway: The ratios oscillate and settle toward the golden ratio $\approx 1.618$ — a preview of limits.