Statistics

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Averages of special series

⏱ 3 min read🧩 5 question types🎯 14 practice Q

The idea in one minute

Some averages can be written down without adding anything: runs of natural numbers, squares, cubes, odds, evens and multiples.

Learn the four sums below. Exams ask for a mean, a total, or the count that hits a given total.

The four working sums

\sum k = \frac{n(n+1)}{2}, \quad \sum k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum k^3 = \left[\frac{n(n+1)}{2}\right]^2

First 10 numbers: sum $= \dfrac{10 \times 11}{2} = 55$ , squares $= 385$ , cubes $= 3025$ .

Rule: The sum of the first n cubes is the square of the sum of the first n numbers: $3025 = 55^2$ .

Means that need no addition

\bar{x} = \frac{\text{sum}}{n}

First n naturals: mean $= \dfrac{n+1}{2}$ , the middle value. First 25 naturals average 13. First n odd numbers: mean $= n$ , because $1 + 3 + \dots + (2n-1) = n^2$ .

First n even numbers: mean $= n + 1$ , since $2 + 4 + \dots + 2n = n(n+1)$ .

Tip: First 9 odds average 9; first 9 evens average 10. One apart, always.

Multiples behave like naturals

The first $n$ multiples of $k$ are $k, 2k, \dots, nk$ : the naturals scaled by $k$ .

\text{sum} = k \cdot \frac{n(n+1)}{2}, \qquad \text{mean} = \frac{k(n+1)}{2}

First 15 multiples of 4: sum $= 4 \times \dfrac{15 \times 16}{2} = 480$ , mean $= 32$ .

Watch: 32 here is not a multiple of 4 belonging to the list. A mean need not be a member of the data.

A slice of the naturals

Numbers from 10 to 20 are just the naturals with the first nine removed.

\text{sum} = \sum_{1}^{q} - \sum_{1}^{p-1}, \qquad \text{count} = q - p + 1

Sum $= 210 - 45 = 165$ over 11 numbers, mean $= \dfrac{165}{11} = 15$ . The shortcut agrees: an equally spaced run averages its ends, $\dfrac{10+20}{2} = 15$ .

Consecutive numbers in a row

Consecutive numbers sit evenly around their middle. The mean equals the middle term: one middle for odd counts, half-way between two middles for even counts.

Five consecutive even numbers average 16, so the middle one is 16: the list is 12, 14, 16, 18, 20 and the largest is 20. For any equally spaced list, largest $=$ mean $+$ (span off the middle).

Example: Seven consecutive integers averaging 25 run from 22 to 28; no adding needed.

Working backwards from a total

Give the count a letter, write the sum formula, and solve.

The sum of the first n cubes is 2025:

\left[\frac{n(n+1)}{2}\right]^2 = 2025 = 45^2 \Rightarrow \frac{n(n+1)}{2} = 45 \Rightarrow n = 9

Tip: Spot perfect squares. $\sum k^3$ is always a perfect square, so the square root hands you $\sum k$ .

Mean of squares and cubes

For the mean, divide the sum by n.

Mean of the first 10 squares $= \dfrac{385}{10} = 38.5$ . Mean of the first 10 cubes $= \dfrac{3025}{10} = 302.5$ .

Watch: These means sit well above the mean of the numbers themselves (5.5), because big values grow fast when squared or cubed.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1common3 practice Q

Mean of squares or cubes

How to spot it:

The average of the first n squares or cubes is asked.

\bar{x} = \frac{1}{n}\sum k^r

Method

Write the matching sum formula.
Substitute n.
Divide by n for the mean.

Why it works:

The sums have closed forms, so the mean is one substitution away.

Try this

Find the mean of the squares of the first 10 natural numbers.

Show solution

$\sum k^2 = \dfrac{10 \times 11 \times 21}{6} = 385$ .
Mean $= 385 \div 10 = 38.5$ .

Answer

38.5

Type 2very common4 practice Q

Consecutive numbers and the middle term

How to spot it:

The average of consecutive integers/odds/evens is given; an end value is asked.

\text{mean} = \text{middle term}

Method

Set the middle term equal to the average.
Step out to the wanted end.
Odd count: one middle; even count: two middles.

Why it works:

Equal spacing makes the average sit exactly at the middle of the row.

Try this

The average of five consecutive even numbers is 16. The largest of them is:

Show solution

Middle term $= 16$ .
List: 12, 14, 16, 18, 20.
Largest $= 20$ .

Answer

Type 3common2 practice Q

A total that fixes the count

How to spot it:

A sum of naturals, squares or cubes is given; find how many terms make it.

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

Method

Match the total to its family (perfect square points to cubes).
Write the sum formula equal to the total.
Solve the small equation for n.

Why it works:

Each sum formula grows strictly with n, so one n fits one total.

Try this

The sum of the cubes of the first n natural numbers is 2025. Find n.

Show solution

$\sum k^3 = \left[\dfrac{n(n+1)}{2}\right]^2 = 2025 = 45^2$ .
$\dfrac{n(n+1)}{2} = 45$ .
$n(n+1) = 90 \Rightarrow n = 9$ .

Answer

n = 9

Type 4common2 practice Q

Means of odds and evens

How to spot it:

The average of the first n odd or even numbers is asked.

\text{odds: } \bar{x} = n; \quad \text{evens: } \bar{x} = n+1

Method

Count how many terms there are.
Odds: the mean is that count.
Evens: the mean is one more.

Why it works:

The kth odd is $2k-1$ and the kth even is $2k$ ; their averages differ by one.

Try this

Find the average of the first 9 odd numbers.

Show solution

Sum $= 9^2 = 81$ .
Mean $= 81 \div 9 = 9$ .

Answer

Type 5occasional

Multiples of a number

How to spot it:

The mean or total of the first n multiples of k is asked.

\text{mean} = \frac{k(n+1)}{2}

Method

Confirm the list: $k, 2k, \dots, nk$ .
Mean: $k$ times the naturals' mean.
Total: mean times n.

Why it works:

Multiples of k are the first n naturals scaled by k.

Try this

Find the mean of the first 15 multiples of 4.

Show solution

Mean $= 4 \times \dfrac{15 + 1}{2} = 4 \times 8$ .
$= 32$ .

Answer

Formula sheet

Sum of first n naturals

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

Sum of squares

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

Sum of cubes

\sum k^3 = \left[\frac{n(n+1)}{2}\right]^2

The square of the sum of the first n naturals.

Mean of first n naturals

\frac{n+1}{2}

Mean of first n odds

n

Mean of first n evens

n+1

Multiples of k

\text{sum} = \frac{kn(n+1)}{2},\; \text{mean} = \frac{k(n+1)}{2}

Shortcuts that save time

⚡ The middle is the mean

For any equally spaced list the mean is the middle term. Use it forwards (find the mean) and backwards (rebuild the list).

Example

Nine consecutive integers have mean 41. Find the smallest.

Show solution

Middle (5th) term $= 41$ .
Smallest $= 41 - 4 = 37$ .

Answer

⚡ Odds add to squares

1 + 3 + ... up to n odd numbers is exactly n². Use it to test counts fast.

Example

How many consecutive odd numbers starting from 1 add up to 121?

Show solution

$n^2 = 121$ .
$n = 11$ .

Answer

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Averaging the first and last term of a list that is not equally spaced.

The middle-is-the-mean rule needs equal gaps.

Mistake 02

Using the naturals' mean (n+1)/2 for odds or evens.

First n odds average n; first n evens average n+1.

Mistake 03

Forgetting to divide the sum by n when the mean is asked.

Sum formulas give totals; means divide by n.

Mistake 04

Treating the sum of cubes as n(n+1)/2 squared being n+1 squared.

Square the whole triangular sum, not n+1 alone.

Mistake 05

Counting the first n multiples of k as kn numbers.

There are exactly n of them: k, 2k, ..., nk.

Quick revision

Read this the night before the exam.

$\sum k = \dfrac{n(n+1)}{2}$ ; mean of naturals $= \dfrac{n+1}{2}$ .
$\sum k^2 = \dfrac{n(n+1)(2n+1)}{6}$ .
$\sum k^3 = \left(\sum k\right)^2$ .
First $n$ odds: sum $n^2$ , mean $n$ .
First $n$ evens: sum $n(n+1)$ , mean $n+1$ .
First $n$ multiples of $k$ : mean $\dfrac{k(n+1)}{2}$ .
Equally spaced list: mean = middle term.

Practice: 14 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 6 min · wrong answers go to your mistake notebook automatically.

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