Sequences & Progressions

medium importance~1 Q in Tier 123 formulas⚡ 15 shortcuts5 subtopics

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Special Series & Standard Sums

⏱ 3 min read🧩 6 question types🎯 13 practice Q

The idea in one minute

Some sums are so common that formulas exist: $\sum n = \dfrac{n(n+1)}{2}$ , $\sum n^2 = \dfrac{n(n+1)(2n+1)}{6}$ , $\sum n^3 = \left[\dfrac{n(n+1)}{2}\right]^2$ . Telescoping fractions cancel in a line, leaving only the ends. One formula pick usually finishes the question.

Three power sums

Three formulas cover most direct asks. For the first $n$ natural numbers:

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

$1 + 2 + \cdots + 40 = \dfrac{40 \times 41}{2} = 820$ .

$1^2 + 2^2 + \cdots + 10^2 = \dfrac{10 \times 11 \times 21}{6} = 385$ .

$1^3 + 2^3 + \cdots + 5^3 = \left(\dfrac{5 \times 6}{2}\right)^2 = 15^2 = 225$ .

Rule: The cube sum is the square of the natural-number sum. Memorise all three shapes.

Odd and even number sums

The first $n$ odd numbers add to a perfect square:

1 + 3 + 5 + \cdots + (2n-1) = n^2

$1 + 3 + \cdots + 25$ has 13 odd numbers, so the sum is $13^2 = 169$ .

The first $n$ even numbers add to $n(n+1)$ : $2 + 4 + \cdots + 20 = 10 \times 11 = 110$ .

An odd-number run that starts later is the difference of two squares. $11 + 13 + \cdots + 29 = 15^2 - 5^2 = 200$ , because 11 is the 6th odd number and 29 the 15th.

Tip: $n^2$ works only when the run starts at 1. Otherwise subtract two square sums.

Count the terms first

From $a$ to $b$ , both included, the count is $b - a + 1$ .

$21 + 22 + \cdots + 50 = \dfrac{50 \times 51}{2} - \dfrac{20 \times 21}{2} = 1275 - 210 = 1065$ .

Subtracting the sum up to 20 removes the first 20 terms cleanly.

Watch: 21 to 50 is 30 terms, not 29. The $+1$ matters.

Telescoping sums

A telescoping sum hides cancellations. Each fraction splits into two parts:

\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}

So $\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \cdots + \dfrac{1}{9 \cdot 10} = 1 - \dfrac{1}{10} = \dfrac{9}{10}$ .

Everything in the middle cancels. Only the first and last pieces survive.

The gaps need not be 1. $\dfrac{1}{1 \cdot 5} + \dfrac{1}{5 \cdot 9} + \cdots + \dfrac{1}{37 \cdot 41} = \dfrac{1}{4}\left(1 - \dfrac{1}{41}\right) = \dfrac{10}{41}$ .

Rule: Split each term as a difference of two simple fractions, then cancel in one line.

Sums of consecutive products

$1 \times 2 + 2 \times 3 + \cdots + n(n+1)$ has a ready answer:

\sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(n+2)}{3}

Up to $10 \times 11$ : $\dfrac{10 \times 11 \times 12}{3} = 440$ .

In reverse, a total of 168 means $n(n+1)(n+2) = 504 = 7 \times 8 \times 9$ , so $n = 7$ .

Note: The formula comes from $k(k+1) = \dfrac{k^3 - k}{3}$ plus the power sums.

Choosing the formula

Read the question, name the last term, then pick:

A plain list of numbers gives $\dfrac{n(n+1)}{2}$ , or a difference of two such sums.
Squares, cubes or products point to the matching formula above.
A chain of fractions points to telescoping.

A sum like $2^2 + 4^2 + \cdots + 20^2$ is $4(1^2 + \cdots + 10^2) = 4 \times 385 = 1540$ .

Tip: Factor out the common piece first; the power sum then applies to small numbers.

Question types you will see

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

Type 1very common3 practice Q

Sum of a run of natural numbers

How to spot it:

A run like 21 + 22 + ... + 50, or a reverse ask: 'the sum 1 to n is S, find n'.

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

Method

Find the count: last minus first plus 1.
Use $\dfrac{n(n+1)}{2}$ for each full run.
Subtract the part before the start.

Why it works:

A run from $a$ to $b$ is (sum up to $b$ ) minus (sum up to $a - 1$ ).

Try this

Find the sum 21 + 22 + ... + 50.

Show solution

$\dfrac{50 \times 51}{2} - \dfrac{20 \times 21}{2}$ .
$1275 - 210 = 1065$ .

Answer

1065

Type 2common2 practice Q

Sum of squares

How to spot it:

A list of squares 1² + 2² + ... + n², sometimes with a factor like even squares only.

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

Method

Read $n$ from the last square.
Apply the formula.
Factor out common pieces first for even or odd squares.

Why it works:

The closed form replaces adding every square individually — one formula, no listing.

Try this

Find the sum 1² + 2² + 3² + ... + 10².

Show solution

$\dfrac{10 \times 11 \times 21}{6}$ .
$= 385$ .

Answer

385

Type 3occasional2 practice Q

Sum of cubes

How to spot it:

1³ + 2³ + ... appears, or a cube sum equals a square in disguise.

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

Method

Compute $\dfrac{n(n+1)}{2}$ .
Square it.

Why it works:

The cube sum equals the square of the triangular number — the fastest cube fact in exams.

Try this

Find the sum 1³ + 2³ + 3³ + 4³ + 5³.

Show solution

$\dfrac{5 \times 6}{2} = 15$ .
$15^2 = 225$ .

Answer

225

Type 4common2 practice Q

Sum of odd and even numbers

How to spot it:

Runs of odd or even numbers, starting at 1 (or 2) or later.

1 + 3 + \cdots + (2n-1) = n^2

Method

Count the terms in the run.
Start at 1: the sum is the count squared.
Start later: subtract two square sums.

Why it works:

Each new odd number extends the previous square by one L-shaped layer, so $n$ odds make $n^2$ .

Try this

Find the sum 11 + 13 + 15 + ... + 29.

Show solution

$= (1 + \cdots + 29) - (1 + \cdots + 9)$ .
$15^2 - 5^2 = 200$ .

Answer

200

Type 5common2 practice Q

Telescoping fraction sums

How to spot it:

A fraction chain with overlapping denominators: 1/(1·2) + 1/(2·3) + ..., or gaps of more than 1.

\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}

Method

Split each term into a difference.
Cancel the diagonal pairs.
Keep the first piece of term 1 and the last piece of the final term.

Why it works:

Each split term cancels most of the next one, so the whole chain collapses to its two ends.

Try this

Find the sum 1/(1·2) + 1/(2·3) + ... + 1/(9·10).

Show solution

Split each term.
Sum $= 1 - \dfrac{1}{10}$ .

Answer

9/10

Type 6occasional2 practice Q

Sums of consecutive products

How to spot it:

1×2 + 2×3 + 3×4 + ... — neighbouring whole numbers multiply.

\sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(n+2)}{3}

Method

Read $n$ from the last product.
Apply $\dfrac{n(n+1)(n+2)}{3}$ .
Reverse asks: match the total with three consecutive numbers.

Why it works:

The product chain equals one third of $n(n+1)(n+2)$ , so both directions are one line.

Try this

Find the sum 1×2 + 2×3 + 3×4 + ... + 10×11.

Show solution

$\dfrac{10 \times 11 \times 12}{3}$ .
$= 440$ .

Answer

440

Formula sheet

Sum of first n numbers

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

Sum of first n squares

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

Sum of first n cubes

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

The square of the natural-number sum.

First n odd numbers

1 + 3 + \cdots + (2n-1) = n^2

The run must start at 1.

First n even numbers

2 + 4 + \cdots + 2n = n(n+1)

Telescoping split

\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}

Consecutive products

\sum_{k=1}^{n} k(k+1) = \frac{n(n+1)(n+2)}{3}

Shortcuts that save time

⚡ Pair the ends of a natural-number sum

1 + 40, 2 + 39, ... every pair adds to 41. With 20 pairs, the sum is $20 \times 41$ . The formula $\dfrac{n(n+1)}{2}$ does the same in one line.

Example

Find the sum 1 + 2 + 3 + ... + 40.

Show solution

$\dfrac{40 \times 41}{2}$ .
$= 820$ .

Answer

820

⚡ Odd numbers build squares

Count the odd numbers; the sum is that count squared. A run that starts past 1 is a difference of two squares.

Example

Find the sum 1 + 3 + 5 + ... + 25.

Show solution

Odd numbers: $\dfrac{25+1}{2} = 13$ .
Sum $= 13^2 = 169$ .

Answer

169

⚡ Telescoping: keep only the ends

Split each fraction as a difference. Writing the first few terms shows the middle cancelling in a diagonal line.

Example

Find the sum 1/(1·2) + 1/(2·3) + ... + 1/(9·10).

Show solution

$\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1}$ .
Sum $= 1 - \dfrac{1}{10} = \dfrac{9}{10}$ .

Answer

9/10

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $\dfrac{n(n+1)}{2}$ for squares or cubes — each power has its own formula.

Mistake 02

Forgetting to square $\left[\dfrac{n(n+1)}{2}\right]$ in the sum of cubes.

Mistake 03

Counting terms as $b - a$ — from $a$ to $b$ inclusive there are $b - a + 1$ terms.

Mistake 04

Squaring the count for odd numbers that do not start at 1 — $11 + 13 + \cdots + 29$ is $15^2 - 5^2$ , not $10^2$ .

Mistake 05

Adding telescoping fractions one by one — pair the cancellations; only the two ends survive.

Quick revision

Read this the night before the exam.

$\sum n = \dfrac{n(n+1)}{2}$ , $\sum n^2 = \dfrac{n(n+1)(2n+1)}{6}$ , $\sum n^3 = \left[\dfrac{n(n+1)}{2}\right]^2$ .
First $n$ odd numbers: $n^2$ ; first $n$ even numbers: $n(n+1)$ .
Run from $a$ to $b$ : subtract two power sums; count is $b - a + 1$ .
Telescoping: $\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1}$ ; only the ends survive.
$1 \times 2 + \cdots + n(n+1) = \dfrac{n(n+1)(n+2)}{3}$ .
Factor common pieces (like $2^2$ ) before applying a power sum.

Practice: 13 questions

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

Topic test · 13 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.

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