Identity

Square of a sum of several terms

Algebra • Expressions • Algebraic Identities

Formula

(\sum_{i} a_{i})^{2} = \sum_{i} a_{i}^{2} + 2 \sum_{i < j} a_{i} a_{j}

For any numbers

a_{1}, a_{2}, \dots, a_{n}

(a_{1} + a_{2} + \dots + a_{n})^{2} = a_{1}^{2} + a_{2}^{2} + \dots + a_{n}^{2} + 2 \sum_{i < j} a_{i} a_{j}

; for three terms this reads

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)

How to use it

1Rearranged one way it gives the sum of squares: $\sum a_{i}^{2} = (\sum a_{i})^{2} - 2 \sum_{i < j} a_{i} a_{j}$ .
2Rearranged the other way it gives the sum of products two at a time: $\sum_{i < j} a_{i} a_{j} = \frac{1}{2} [(\sum a_{i})^{2} - \sum a_{i}^{2}]$ .
3When the terms add to zero the first bracket vanishes, which ties the sum of squares and the sum of pairwise products together.

Worked example

Given

a + b + c = 9

and

ab + b c + c a = 23

, find

a^{2} + b^{2} + c^{2}

Step-by-step

1
$a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2 (ab + b c + c a)$
Why: Rearrange the three-term case of the identity.
2
$a^{2} + b^{2} + c^{2} = 81 - 46 = 35$

Answer

35

Watch out

Each cross product appears once in

\sum_{i < j}

and the identity supplies the doubling; counting

a_{i} a_{j}

and

a_{j} a_{i}

separately doubles it a second time. The familiar two-term case

(a + b)^{2} = a^{2} + 2 ab + b^{2}

is this same identity, not a different one.

Builds on

Square of a sum and of a difference

Also known as

square of a sum of n terms, (a+b+c)^2, sum of products two at a time, elementary symmetric sum, sum of squares from sum and pair sum, square of a trinomial

Appears in past papers

Formula

Worked example

Appears in past papers

2020 (II)

2019 (I)

2018 (II)