Sequences and Series

A sequence is an ordered list of terms, which may be finite or infinite. A series is the result of adding terms from a sequence. The word progression is often used as a general label for a sequence or its associated series. The position of a term matters: $u_1$ is the first term, $u_2$ is the second term, and $u_n$ is the term in position $n$.

An $n$th-term formula gives a term directly from its position. For example, if

$$u_n = 4n-3,$$

then substituting $n=1,2,3,\ldots$ gives the sequence

$$1,5,9,13,\ldots$$

The subscript is a label, not multiplication: $u_n$ means the term in position $n$.

An explicit formula can be used in two directions:

• substitute a position to find the corresponding term;
• set the formula equal to a given value and solve for the position.

Worked example

Let

$$u_n=3n^2-2.$$

The first three terms are

$$u_1=1,\qquad u_2=10,\qquad u_3=25.$$

To determine whether $73$ is a term, solve

$$3n^2-2=73.$$

This gives $n^2=25$, so $n=5$ because a term number is a positive integer. Therefore, $73$ is the fifth term.

A finite list of terms does not usually determine a unique formula. Many different rules can begin with the same few values, so a claimed pattern must be supported by the rule stated in the question rather than by guesswork alone.

The notation

$$\sum_{k=p}^{q}u_k$$

means

$$u_p+u_{p+1}+\cdots+u_q.$$

There are $q-p+1$ terms because both endpoints are included. This inclusive count is a frequent source of errors.

TMUA strategy: Write out two or three terms whenever an indexed expression looks complicated. This often exposes an incorrect starting index, final index or exponent before substantial algebra is done.

A recurrence relation defines each new term using an earlier term. A simple recurrence has the form

$$x_{n+1}=f(x_n),$$

together with an initial value such as $x_1=a$. The initial value is essential: without it, the recurrence usually describes many possible sequences.

To generate the sequence, substitute the current term into the rule repeatedly.

Worked example

Suppose

$$x_1=2,\qquad x_{n+1}=|x_n-3|.$$

Then

$$x_2=|2-3|=1,$$

$$x_3=|1-3|=2,$$

and the values continue as

$$2,1,2,1,\ldots$$

The sequence alternates with period $2$.

For a recurrence depending only on the immediately preceding term, once a value repeats, all later terms repeat in the same way. In the example, $x_3=x_1$, so applying the same function again must produce $x_4=x_2$, then $x_5=x_3$, and so on.

A value $L$ satisfying

$$L=f(L)$$

is called a fixed value of the recurrence. If some term equals $L$, every subsequent term also equals $L$. For example, if $x_{n+1}=\frac{x_n+6}{2}$, then $L=6$ is fixed because

$$6=\frac{6+6}{2}.$$

You are expected to generate and reason about simple recurrences, but not to solve general recurrence relations by advanced methods.

Common traps

• Do not replace $x_n$ with $n$. The input to $f$ is the previous term, not the term number.
• Check the starting index. A rule beginning with $x_0$ produces different labels from one beginning with $x_1$.
• Do not claim a cycle merely because the first few terms look repetitive. For a one-step recurrence, a cycle is guaranteed once an earlier value actually reappears.

An arithmetic sequence changes by a constant amount each time. This constant is the common difference $d$:

$$u_{n+1}-u_n=d.$$

If the first term is $a$, then moving from the first term to the $n$th term requires $n-1$ steps. Therefore,

$$u_n=a+(n-1)d.$$

The $n-1$ is important: the first term has had no common differences added.

For the sum of the first $n$ terms, write the sum forwards and backwards:

$$S_n=a+(a+d)+\cdots +(a+(n-1)d),$$

$$S_n=(a+(n-1)d)+(a+(n-2)d)+\cdots+a.$$

Adding corresponding terms gives $n$ identical pairs, each equal to $2a+(n-1)d$. Hence

$$2S_n=n\bigl(2a+(n-1)d\bigr),$$

so

$$S_n=\frac{n}{2}\bigl(2a+(n-1)d\bigr).$$

If the last term is $l$, this can be written more simply as

$$S_n=\frac{n}{2}(a+l).$$

This says that the sum is

$$\text{number of terms}\times\text{average of the first and last terms}.$$

The natural numbers $1,2,3,\ldots,n$ form an arithmetic sequence with $a=1$, $d=1$ and last term $n$. Therefore,

$$1+2+\cdots+n=\frac{n(n+1)}{2}.$$

Worked example

An arithmetic sequence begins

$$7,11,15,\ldots$$

Here $a=7$ and $d=4$. Its 25th term is

$$u_{25}=7+24(4)=103.$$

The sum of its first 25 terms is

$$S_{25}=\frac{25}{2}(7+103)=1375.$$

For a consecutive block from the $p$th term to the $q$th term, the number of terms is $q-p+1$. You may either treat the block as a new arithmetic sequence or subtract partial sums:

$$u_p+u_{p+1}+\cdots+u_q=S_q-S_{p-1}.$$

Efficient TMUA methods

• When the first and last terms are easy to identify, use $S_n=\frac{n}{2}(a+l)$ rather than calculating $d$ unnecessarily.
• To test whether a sequence is arithmetic, compare consecutive differences rather than relying on appearance.
• If the question asks only for a sum, look for symmetry: terms equally far from the ends have the same total.

Common mistakes

• Using $a+nd$ instead of $a+(n-1)d$.
• Confusing the $n$th term $u_n$ with the sum $S_n$.
• Counting the terms from position $p$ to position $q$ as $q-p$ instead of $q-p+1$.

A geometric sequence is formed by multiplying by a constant common ratio $r$:

$$\frac{u_{n+1}}{u_n}=r$$

whenever the ratio is defined. If the first term is $a$, then

$$u_n=ar^{n-1}.$$

The sum of the first $n$ terms is

$$S_n=a+ar+ar^2+\cdots+ar^{n-1}.$$

Multiplying by $r$ gives

$$rS_n=ar+ar^2+\cdots+ar^{n-1}+ar^n.$$

Subtracting the second equation from the first causes the middle terms to cancel:

$$S_n-rS_n=a-ar^n.$$

Therefore, for $r\ne1$,

$$S_n=\frac{a(1-r^n)}{1-r}.$$

The equivalent form

$$S_n=\frac{a(r^n-1)}{r-1}$$

is sometimes more convenient when $r>1$. If $r=1$, every term equals $a$, so

$$S_n=na.$$

Worked example

Find the sum of the first eight terms of

$$5,10,20,40,\ldots$$

Here $a=5$, $r=2$ and $n=8$. Thus

$$S_8=\frac{5(1-2^8)}{1-2}=1275.$$

For the block

$$ar^m+ar^{m+1}+\cdots+ar^n,$$

the first term is $ar^m$, the common ratio is $r$, and the number of terms is $n-m+1$. Hence

$$ar^m+ar^{m+1}+\cdots+ar^n =ar^m\frac{1-r^{n-m+1}}{1-r}.$$

It can also be viewed as a difference of partial sums, but the indices must be handled carefully:

$$ar^m+ar^{m+1}+\cdots+ar^n=S_{n+1}-S_m,$$

where $S_k=a+ar+\cdots+ar^{k-1}$.

Common traps

• The common ratio may be negative, producing alternating signs.
• In $ar^{n-1}$, the first term corresponds to exponent $0$.
• Do not use the finite-sum formula with $r=1$ because it would divide by zero.
• Check whether the formula uses $1-r^n$ over $1-r$ or $r^n-1$ over $r-1$; mixing the two forms changes the sign.

An infinite series is interpreted through its partial sums. If $S_n$ is the sum of the first $n$ terms and the values of $S_n$ approach a fixed number as $n$ increases, the series converges to that number. A necessary first check is that the individual terms approach $0$. For a geometric series,

$$a+ar+ar^2+\cdots,$$

this happens precisely when

$$|r|<1.$$

Under this condition, $r^n\to0$ as $n\to\infty$. Starting from

$$S_n=\frac{a(1-r^n)}{1-r},$$

we obtain

$$S_\infty=\frac{a}{1-r}.$$

The condition is on the magnitude of $r$, not merely on whether $r<1$. For example, $r=-\frac12$ converges, but $r=-2$ does not.

If $|r|\ge1$ and $a\ne0$, the geometric series does not converge:

• $r=1$ gives repeated non-zero terms;
• $r=-1$ makes the partial sums oscillate;
• $|r|>1$ makes the term magnitudes grow;
• for other boundary cases, the terms fail to approach zero.

Worked example

Consider

$$12-6+3-\frac32+\cdots$$

The first term is $12$ and the common ratio is $-\frac12$. Since

$$\left|-\frac12\right|<1,$$

the series converges and

$$S_\infty=\frac{12}{1-(-1/2)}=8.$$

The sum from the $m$th term onwards is itself an infinite geometric series whose first term is $u_m$. Therefore,

$$u_m+u_{m+1}+u_{m+2}+\cdots=\frac{u_m}{1-r}$$

when $|r|<1$.

TMUA strategy: Before using an infinite-sum formula, state or check $|r|<1$. This takes little time and prevents one of the most common errors in this topic.

Do not confuse the terms of a sequence with the partial sums of its series. In a convergent geometric series, the individual terms approach $0$, while the partial sums approach $\frac{a}{1-r}$.

For a non-negative integer $n$, factorial notation is defined by

$$n!=n(n-1)(n-2)\cdots 2\cdot1,$$

with

$$0!=1.$$

For example,

$$5!=5\cdot4\cdot3\cdot2\cdot1=120.$$

The binomial coefficient

$$\binom{n}{r}$$

counts the number of ways to choose $r$ objects from $n$ objects when order does not matter. For integers satisfying $0\le r\le n$,

$$\binom{n}{r}=\frac{n!}{r!(n-r)!}.$$

For example,

$$\binom{10}{3}=\frac{10!}{3!7!} =\frac{10\cdot9\cdot8}{3\cdot2\cdot1} =120.$$

Cancelling before evaluating the factorials is usually faster and reduces arithmetic errors.

The coefficients have useful symmetry:

$$\binom{n}{r}=\binom{n}{n-r}.$$

Choosing the $r$ objects to keep is equivalent to choosing the $n-r$ objects to leave out. In particular,

$$\binom{n}{0}=\binom{n}{n}=1.$$

Pascal's triangle arranges the binomial coefficients so that each interior entry is the sum of the two entries above it. This gives

$$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}.$$

Pascal's triangle is efficient for small powers. For a large power or a question asking for only one term, the factorial formula is usually quicker.

For a positive integer $n$, the binomial theorem expands a power of two terms:

$$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r.$$

Written out,

$$(a+b)^n =a^n+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2+\cdots+b^n.$$

For $(1+x)^n$, this becomes

$$(1+x)^n =1+\binom{n}{1}x+\binom{n}{2}x^2+\cdots+x^n.$$

The general term is especially useful. The term involving $b^r$ is

$$T_{r+1}=\binom{n}{r}a^{n-r}b^r.$$

It is called the $(r+1)$th term because $r=0$ gives the first term.

For an expression of the form $(a+f(x))^n$,

$$T_{r+1}=\binom{n}{r}a^{n-r}[f(x)]^r.$$

Keep the whole of $f(x)$ in brackets before raising it to a power. This preserves constants and signs correctly.

Worked example: finding a coefficient

Find the coefficient of $x^4$ in

$$(2-3x)^6.$$

The $x^4$ term occurs when $r=4$:

$$\binom{6}{4}2^{6-4}(-3x)^4.$$

Therefore, the coefficient is

$$\binom{6}{4}2^2(-3)^4 =15\cdot4\cdot81 =4860.$$

The sign is positive because the negative term is raised to an even power.

Worked example: finding a constant term

Find the constant term in

$$\left(x^{-2}+2x\right)^6.$$

The general term is

$$\binom{6}{r}(x^{-2})^{6-r}(2x)^r.$$

Its power of $x$ is

$$-2(6-r)+r=-12+3r.$$

A constant term has power $0$, so

$$-12+3r=0,$$

which gives $r=4$. The constant term is therefore

$$\binom{6}{4}2^4=240.$$

Efficient TMUA methods

• When asked for one term or coefficient, use the general term instead of expanding everything.
• Determine the required value of $r$ from the power of $x$ before doing numerical arithmetic.
• Use $\binom{n}{r}=\binom{n}{n-r}$ when it makes the calculation smaller.
• For consecutive terms, update systematically: the power of the first binomial part decreases by $1$, while the power of the second increases by $1$.

Common mistakes

• Forgetting that in $(-3x)^r$, the minus sign and the $3$ are both raised to the power $r$.
• Giving an entire term when the question asks only for its coefficient.
• Confusing the index $r$ with the term number: $r=4$ gives the fifth term.
• Using the positive-integer binomial expansion when the power is negative or fractional. Such expansions are not part of this topic.
• Assuming every requested power of $x$ must occur. If the equation for $r$ has no integer solution with $0\le r\le n$, that power is absent and its coefficient is $0$.