Methods in Sequences and Summations

What you will get from this section. You will see a collection of useful examples involving arithmetic and geometric sequences, recurrence relations and summations. Each example develops a technique for exposing a familiar sequence hidden inside a less familiar problem.

TMUA Relevance Score: 8/10

I am writing this article because sequence and summation questions give examiners plenty of opportunities to set some genuinely interesting traps—and then sit back with popcorn and watch you fall into them. 😂

There is no single grand result tying everything in this article together. However, exposure to all these different types of problem provides exactly the kind of experience that is usually gained only after attempting many, many questions. Here, I am laying many of those traps out plainly for you to inspect, so examiners will have a much harder time catching you out with them later. Good, right?! 😁

Arithmetic and geometric sequences

I will assume that you are already familiar with arithmetic and geometric sequences, recurrence relations and the standard formulae, so these will not be our focus. Instead, the examples that follow explore some less well-known results and useful techniques involving these sequences.

Example 1: Arithmetic–geometric sequence

For a real number $x$, the sequence $(u_n)$ is defined by $u_n=(2n-1)x^{n-1}$. Find $x$ such that

Example 1 solution

The factor $2n-1$ generates the arithmetic sequence $1,3,5,7,\ldots$, while $x^{n-1}$ generates the geometric sequence $1,x,x^2,x^3,\ldots$. The terms of $(u_n)$ are formed by multiplying these two sequences term by term.

Let the given sum be $S$. Multiplying it by the common ratio $x$ gives

Subtracting the second expression from the first gives

The terms starting from $2x$ form an infinite geometric series with common ratio $x$. Since the original sum converges, this geometric series must also converge, so we now know that $|x|<1$.

Using the infinite geometric-series formula,

Substituting $S=6$ gives

The two algebraic possibilities are $x=\frac53$ and $x=\frac12$. Since $|x|<1$, we must have $x=\frac12$.

Technique. When the terms are formed by multiplying an arithmetic sequence by a geometric sequence, multiply the entire sum by the geometric common ratio and subtract to expose a new geometric series.

Example 2: Pair analysis in an arithmetic sequence

The numbers $u_1,u_2,\ldots,u_{18}$ are consecutive terms of an arithmetic sequence, and $u_5+u_{14}=40$.

Consider the following three statements.

1. $u_9+u_{10}=40$.
2. $u_1+u_2+\cdots+u_{18}=360$.
3. $u_1+u_4+u_7+u_9+u_{12}+u_{15}+u_{18}=140$.

Which of these statements must be true?

Example 2 solution

For the sum of two terms in an arithmetic sequence, shifting one term down by one index position and the other up by one index position decreases one term by the common difference $d$ and increases the other by $d$. These two effects cancel, so the sum remains unchanged.

Therefore, $u_5+u_{14}=u_4+u_{15}=40$.

Of course, we may shift by more than one index position. Moving each term by two positions changes one by $+2d$ and the other by $-2d$, so the two effects still cancel:

$u_5+u_{14}=u_7+u_{12}=40$.

More generally,

whenever $a+b=p+q$. This is true for every arithmetic sequence. I call this pair analysis—a name I made up myself!

Statement 1: The indices $9$ and $10$ add to $19$, as do the indices $5$ and $14$. Pair analysis therefore gives $u_9+u_{10}=u_5+u_{14}=40$.

Statement 1 must be true.

Statement 2: Pair the terms from the outside in:

$(u_1+u_{18})+(u_2+u_{17})+\cdots+(u_9+u_{10})$.

Each pair has index sum $19$ and therefore has value $40$. There are $9$ pairs, so $u_1+u_2+\cdots+u_{18}=9\times40=360$.

Statement 2 must also be true.

Statement 3: Pair as many terms as possible:

If statement 3 were true, this would force $u_9$ to be exactly $20$.

But hold on! The only condition we were given at the start was $u_5+u_{14}=40$, which is equivalent to $u_9+u_{10}=40$. This clearly cannot determine $u_9$ exactly. If it did, then $u_{10}=40-u_9$ would also be known. We could then find the common difference $d=u_{10}-u_9$, recover the first term and thereby determine the entire arithmetic sequence from just one pair-sum condition—which is clearly impossible! 😂

That is my rather long-winded way of saying that any subsequent claim to have determined $u_9$ exactly is fraudulent! Therefore, statement 3 need not be true.

Hence statements 1 and 2 must be true, while statement 3 need not be true.

Technique. I call this pair analysis. There is a directly analogous version for geometric sequences.

In an arithmetic sequence, $u_a+u_b=u_p+u_q$ whenever $a+b=p+q$.

In a geometric sequence, $u_a\times u_b=u_p\times u_q$ whenever $a+b=p+q$.

Example 3: Averages in arithmetic sequences

A non-constant arithmetic sequence consists entirely of positive integers and has $n$ terms. The sum of all its terms is $105$.

The sum of the terms in odd-numbered positions is $15$ greater than the sum of the terms in even-numbered positions.

Find all possible values of $n$.

Example 3 solution

Let $S_o$ and $S_e$ denote the sums of the terms in odd-numbered and even-numbered positions respectively. We are given that $S_o+S_e=105$ and $S_o-S_e=15$, so $S_o=60$ and $S_e=45$.

We must consider separately whether $n$ is odd or even.

Case 1: $n$ is odd

Let $M$ be the middle term. Then $M$ is the average of the terms in odd-numbered positions and also the average of the terms in even-numbered positions.

Since $n$ is odd, there is exactly one more term in an odd-numbered position than in an even-numbered position. Therefore, $S_o-S_e=M$, giving $M=15$.

The middle term is also the average of the entire sequence, so $105=nM=15n$, giving $n=7$.

Case 2: $n$ is even

There are $n/2$ terms in each group. Every term in an even-numbered position is obtained by adding $d$ to the preceding term in an odd-numbered position, so

Every term is an integer, so $d$ must also be an integer. Therefore, $n$ must be an even divisor of $30$, giving $n=2,6,10$ or $30$.

Now we apply the positivity restriction. Since $d=-30/n<0$, the sequence is decreasing and its last term is its smallest. We can restrict $n$ further with a little more work.

The sum of the first and last terms is twice the average of the sequence. We also know the difference between these terms:

Adding these equations and dividing by $2$ gives

Since $u_n$ must be positive, $\frac{120}{n}-15>0$, giving $n<8$. Therefore, only $n=2$ and $n=6$ remain from the even case.

These values are all genuinely possible: the sequences $60,45$; $30,25,20,15,10,5$; and $12,13,14,15,16,17,18$ provide examples for $n=2,6$ and $7$ respectively.

Therefore, the possible values of $n$ are $2,6$ and $7$.

Technique. The average of an arithmetic sequence is the average of its first and last terms or, more generally, of any pair of terms equally far from the two ends. If the sequence has an odd number of terms, this average is also the middle term.

For the curious. My original version of this question asked for the total number of distinct sequences satisfying these conditions. What do you think the answer is? Hint: it is not $6$. 😂

Sequences defined by recurrence relations

Examples 1–3 were complete problems. The examples below are deliberately much simpler, as they are just here to demonstrate useful techniques. A useful question to keep in mind when dealing with recurrence relations is: Is there a simpler related sequence to consider?

Example 4: A hidden arithmetic sequence

The sequence $(u_n)$ is defined by $u_1=4$, $u_2=7$ and $u_{n+1}+u_{n-1}=2u_n$ for $n\geq2$.

Rearranging the recurrence gives

Successive differences are therefore equal. Since $u_2-u_1=3$, the sequence is arithmetic with common difference $3$, giving $u_n=4+3(n-1)=3n+1$.

Technique. A recurrence involving three consecutive terms may hide an arithmetic or geometric sequence. Look for:

• $u_{n+1}-u_n=u_n-u_{n-1}$, revealing an arithmetic sequence;
• $\frac{u_{n+1}}{u_n}=\frac{u_n}{u_{n-1}}$, revealing a geometric sequence, provided the required divisions are valid.

Example 5: Taking reciprocals

The positive sequence $(u_n)$ is defined by $u_1=\frac12$ and

The fraction becomes much simpler when inverted:

Define $v_n=1/u_n$. Then $v_1=2$ and $v_{n+1}=v_n+3$, so $(v_n)$ is arithmetic. Therefore, $v_n=3n-1$ and hence

Technique. If a recurrence contains $u_n$ in both the numerator and denominator, try taking reciprocals. This may reveal a simpler related sequence.

Example 6: Shifting to create a geometric sequence

The sequence $(u_n)$ is defined by $u_1=1$ and $u_{n+1}=2u_n+3$.

Without the added $3$, the recurrence would define a geometric sequence with common ratio $2$. This suggests looking for a constant $k$ such that $(u_n+k)$ is geometric with common ratio $2$.

We therefore want

Using the recurrence, this becomes

so $k=3$.

Define $v_n=u_n+3$. Then $v_1=4$ and $v_{n+1}=2v_n$, so $(v_n)$ is geometric. Hence $v_n=4\cdot2^{n-1}=2^{n+1}$, giving

Technique. For a recurrence of the form $u_{n+1}=au_n+b$, look for a constant $k$ such that $(u_n+k)$ is geometric by requiring $u_{n+1}+k=a(u_n+k)$.

Example 7: Applying two transformations

The positive sequence $(u_n)$ is defined by $u_1=1$ and

Taking reciprocals and defining $v_n=1/u_n$ gives $v_1=1$ and $v_{n+1}=2v_n+3$.

This first transformation has not yet produced an arithmetic or geometric sequence, but it has exposed the recurrence from Example 6. Adding $3$ gives $v_{n+1}+3=2(v_n+3)$.

Therefore, $(v_n+3)$ is geometric with first term $4$ and common ratio $2$. Hence $v_n+3=2^{n+1}$, so

Technique. One transformation may reveal the need for another!

Example 8: Rescaling by the term number

The sequence $(u_n)$ is defined by $u_1=1$ and

The repeated appearance of $n$ and $n+1$ suggests dividing each term by its index. Dividing the recurrence by $n+1$ gives

Define $v_n=u_n/n$. Then $v_1=1$ and $v_{n+1}=v_n+1$, so $v_n=n$. Consequently, $u_n=nv_n=n^2$.

Thus, the recurrence was simply defining the sequence $1,4,9,16,\ldots$. In a TMUA question, never underestimate the value of calculating the first few terms and looking for a pattern!

Technique. When a recurrence repeatedly contains $n$ and $n+1$, try dividing or multiplying each term by its index. A rescaled sequence such as $v_n=u_n/n$ may be much simpler.

Example 9: A recurrence with period three

The sequence $(u_n)$ is defined by $u_1=a$, where $a\neq0,1$, and

This example demonstrates another possibility for a recurrence-defined sequence: it may be periodic. How might we recognise this? There are a couple of clues:

• Treating the right-hand side as a function, $f(x)=1-\frac1x$ is a composition of the two self-inverse functions $x\mapsto1/x$ and $x\mapsto1-x$. This suggests possible periodic behaviour.

• Try a simple value such as $a=2$. The first four terms are $2,\frac12,-1,2$. As soon as we return to $2$, we know that the sequence will repeat and therefore has period $3$ for this particular starting value.

Neither observation proves that every permitted starting value produces a sequence with period $3$. The first motivates the experiment, while the second confirms only the case $a=2$. For a complete check, calculate

Therefore, $u_{n+3}=u_n$ for every permitted value of $a$, so the period must divide $3$.

Since $3$ is prime, the period can only be $1$ or $3$. A period of $1$ would require $a$ to be a fixed point of $f$. Solving $f(a)=a$ gives $a^2-a+1=0$, whose discriminant is $-3$, so there are no real solutions.

Once the periodic nature is known, high-index terms and long sums can be reduced to calculations involving only one cycle.

Technique. A recurrence built from simple self-inverse functions, such as $x\mapsto a-x$ and $x\mapsto a/x$, may suggest periodic behaviour. Calculate the first few terms to look for a cycle.

Example 10: Two ways to recognise period two

Before we consider this example, it is worth recalling some facts that should be familiar to students who have studied the functions topic in A-level Mathematics:

The graphs of $f$ and its inverse $f^{-1}$, when the inverse exists, are reflections of each other in the line $y=x$. They satisfy $f^{-1}(f(x))=x$ for every $x$ in the domain of $f$.

Consequently, $f=f^{-1}$ if and only if the graph of $f$ is symmetric about the line $y=x$. This provides a useful way to recognise immediately that a given function is its own inverse, without using algebra.

Consider the recurrence

The graph of $f(x)=5-x$ is clearly symmetric about the line $y=x$. Therefore, $f$ is its own inverse, so applying the recurrence twice returns us to the starting value. The sequence consequently has period $1$ or $2$, depending on its starting value. In particular, $u_1=\frac52$ gives a constant sequence with period $1$; every other starting value gives period $2$.

Notice how we reached this conclusion without any algebra, simply by thinking about the graph of $f(x)=5-x$!

Next, consider

Here, the symmetry of the graph is less obvious, so instead define $f(x)=\frac{x+1}{x-1}$ and check that

Therefore, applying the recurrence twice again returns us to the starting value, so the sequence has period $1$ or $2$. The actual period depends on its starting value $u_1=a$.

If $a$ is a fixed point of $f$, then $f(a)=a$, and the sequence is constant, with period $1$. In this case,

has the two solutions $a=1\pm\sqrt2$. These are also the points where the graph of $f$ intersects the line $y=x$. For every other valid starting value $a\neq1$, the sequence has period $2$.

Technique. To check for period $2$, either recognise that the graph of $y=f(x)$ is symmetric about the line $y=x$ or verify algebraically that $f(f(x))=x$. Fixed points give period $1$; every other valid starting value gives period $2$.

The method of differences

The method of differences is a Further Mathematics topic, although its underlying idea is simple and natural enough to arise in a problem-solving question. You may occasionally encounter the term telescoping sum in community-written paper solutions, mathematics-olympiad material or resources from outside the UK. Do not be confused by the different terminology: it refers to the same cancellation phenomenon. In UK mathematics education, the technique is normally called the method of differences, so that is the name I will use here.

This is probably the least directly relevant part of the article for TMUA preparation because the method is outside the specification. Nevertheless, the required rearrangements are often elementary, and the cancellation itself is certainly fair mathematical game.

Example 11: A hidden difference of consecutive squares

Evaluate $\displaystyle\sum_{r=1}^{100}\frac{2r+1}{r^2(r+1)^2}$.

Example 11 solution

The numerator can be written as the difference of two consecutive squares:

$2r+1=(r+1)^2-r^2$.

This is particularly useful because those same two squares appear in the denominator:

$\displaystyle\frac{2r+1}{r^2(r+1)^2}=\frac{(r+1)^2-r^2}{r^2(r+1)^2}=\frac1{r^2}-\frac1{(r+1)^2}$.

Therefore,

Technique. Look for a way to write the $r$th term as $F(r)-F(r+1)$. When the terms are added, each intermediate value cancels with the corresponding value in the next term, leaving only the two end terms.

Example 12: Partial fractions with a gap of two

Evaluate $\displaystyle\sum_{r=1}^{100}\frac1{r(r+2)}$.

Example 12 solution

Partial fractions give

The sum therefore becomes

Because the two fractions are separated by two indices, two positive terms at the beginning and two negative terms at the end remain. Hence

$\displaystyle\sum_{r=1}^{100}\frac1{r(r+2)}=\frac12\left(1+\frac12-\frac1{101}-\frac1{102}\right)=\frac{7625}{10302}$.

Technique. If the term takes the form $F(r)-F(r+k)$, expect $k$ remaining terms at each end.

Example 13: Two cancelling pairs

Evaluate $\displaystyle\sum_{r=1}^{100}\frac{2}{r(r+1)(r+2)}$.

Example 13 solution

The useful form is not immediately visible. Decomposing the fraction gives

$\frac{2}{r(r+1)(r+2)}=\frac1r-\frac2{r+1}+\frac1{r+2}$.

Now rearrange the three terms as two differences:

$\frac1r-\frac2{r+1}+\frac1{r+2}=\left(\frac1r-\frac1{r+1}\right)-\left(\frac1{r+1}-\frac1{r+2}\right)$.

Each bracket generates its own cancellation. Therefore,

Technique. An expression of the form $F(r)-2F(r+1)+F(r+2)$ can be split into two first differences: $[F(r)-F(r+1)]-[F(r+1)-F(r+2)]$.

Example 14: Trial and error in the method of differences

Evaluate

Example 14 solution

The factors $n+1$ and $n+2$ in the denominator, together with the consecutive powers of $2$, suggest trying a difference such as

Why this particular structure? When $n$ is replaced by $n+1$, the first fraction becomes $\frac{2^{n+1}}{n+2}$, exactly matching the second fraction above. This makes cancellation between consecutive terms possible.

Multiplying both fractions by $2$ gives

which has exactly the same structure. It is simply twice our original trial. Thus, the two guesses have the same structure, so trying either one is sufficient.

However,

could never produce the required cancellation. The second fraction is not what the first fraction becomes when $n$ is replaced by $n+1$: the denominator advances by one index, but the exponent advances by two.

Now simplify our original trial:

This is very close to the required summand: we need only reverse the order and multiply by $2$. Therefore,

Hence

Technique. If you suspect that the method of differences may work, use the structure of the summand to make an informed trial, but make sure that your trial is of the form $F(n)-F(n+k)$.

Summary

Arithmetic and geometric sequences: look for useful products, pairings and averages.

Recurrence relations: try shifting, reciprocating or rescaling the terms to reveal a simpler related sequence, and look out for periodic behaviour.

Method of differences: rewrite the summand as $F(r)-F(r+k)$ so that intermediate terms cancel.