AFCAT Numerical Ability · Sequence and Series

Special Series and the Standard Sums

The sums of the first n natural, odd and even numbers, of squares and of cubes, and series that turn into these sums once their terms are split, paired or rewritten.

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Why this matters

The easy questions are the standard sums: learn five results and each is one line, provided the terms are counted correctly. The HARD ones hide a known series inside another, such as a term that is a whole number minus a GP term, or squares with alternating signs, and the work is seeing how to split or pair the terms.

Concept 1 of 2: The standard sums

Five sums come up again and again. Each is an AP or a known pattern worked out once, so learn the result and skip the working.

Definition

Count the terms first: the odd numbers up to $2n - 1$ are $n$ terms, and the even numbers up to $2n$ are $n$ terms. Then read the sum from the table.

Sum	Result	Example
$1 + 2 + 3 + \dots + n$	$\dfrac{n(n + 1)}{2}$	$1 + 2 + \dots + 40 = 820$
$1 + 3 + 5 + \dots + (2n - 1)$	$n^2$	The first 15 odd numbers add up to $225$
$2 + 4 + 6 + \dots + 2n$	$n(n + 1)$	$2 + 4 + \dots + 40 = 420$
$1^2 + 2^2 + \dots + n^2$	$\dfrac{n(n + 1)(2n + 1)}{6}$	$1^2 + 2^2 + \dots + 10^2 = 385$
$1^3 + 2^3 + \dots + n^3$	$\left[\dfrac{n(n + 1)}{2}\right]^2$	$1^3 + 2^3 + \dots + 10^3 = 3025$ The sum of the cubes is the square of the sum of the numbers.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Sequence and Series · Special Series and Standard Sums

Find the sum of odd numbers from

1

99

Count the odd numbers

The odd numbers from 1 to 79 are 40 numbers, not 79 and not 39, so their sum is

40^2 = 1600

Even numbers are not n squared

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

, one

n

more than the odd-number sum

n^2

. The first 10 even numbers add up to 110, not 100.

Drill 3 more on the standard sums

Concept 2 of 2: Series built from known pieces

A series that is neither an AP nor a GP is often two of them added together, or a standard sum in disguise. Write the kth term, split it, and sum each part with a formula you already know.

Definition

Split the term: if the kth term is a constant minus a GP term, add up the constants and the GP separately.
Pair the terms: with alternating signs, group neighbours. $3^2 - 4^2 = -(3 + 4)$ , because $a^2 - b^2 = (a - b)(a + b)$ .
Find the kth term first: when the terms come from a sum formula, get $a_k = S_k - S_{k-1}$ (Arithmetic Progression page), then sum its square with the standard sums.
Sums split over addition: $\sum (pk + q) = p\sum k + qn$ .

Split and sum

\sum_{k=1}^{n} (u_k + v_k) = \sum_{k=1}^{n} u_k + \sum_{k=1}^{n} v_k

Worked example

Find the sum of the first

n

terms of

0.9 + 0.99 + 0.999 + \dots

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 2 · Sequence and Series · Special Series and Standard Sums

The sum of the first

n

terms of the series

\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots

Pairs need an even count

Pairing

1^2 - 2^2 + 3^2 - \dots

works cleanly only for an even number of terms. With an odd number the last square is left over and has to be added back.

A GP of n terms, not an infinite one

n - (0.1 + 0.01 + \dots)

the bracket stops after

n

terms, so it is

\tfrac 19 (1 - 10^{-n})

. Using the sum to infinity,

\tfrac 19

, is only right when

n

is huge.

Drill 2 more on series built from known pieces

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Series built from known pieces
Split and sum
$\sum_{k=1}^{n} (u_k + v_k) = \sum_{k=1}^{n} u_k + \sum_{k=1}^{n} v_k$

Reference tables (1)

The standard sums5 rows

Sum	Result	Example
$1 + 2 + 3 + \dots + n$	$\dfrac{n(n + 1)}{2}$	$1 + 2 + \dots + 40 = 820$
$1 + 3 + 5 + \dots + (2n - 1)$	$n^2$	The first 15 odd numbers add up to $225$
$2 + 4 + 6 + \dots + 2n$	$n(n + 1)$	$2 + 4 + \dots + 40 = 420$
$1^2 + 2^2 + \dots + n^2$	$\dfrac{n(n + 1)(2n + 1)}{6}$	$1^2 + 2^2 + \dots + 10^2 = 385$
$1^3 + 2^3 + \dots + n^3$	$\left[\dfrac{n(n + 1)}{2}\right]^2$	$1^3 + 2^3 + \dots + 10^3 = 3025$ The sum of the cubes is the square of the sum of the numbers.

Watch out for (4)

Count the odd numbers
→ The standard sums
Even numbers are not n squared
→ The standard sums
Pairs need an even count
→ Series built from known pieces
A GP of n terms, not an infinite one
→ Series built from known pieces

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