Summing Consecutive Numbers

Summing Consecutive Numbers — Moosa Academy

Adding $1 + 2 + 3 + \cdots + 200$ by hand would take all afternoon. One formula does it in a single line — and with two small adjustments the same formula handles sums that start somewhere other than $1$ , or count in tens.

Theorem The consecutive sum formula

$1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2}$

The reason is worth a moment. Pair the numbers from $1$ to $10$ from the outside in:

1 + 10 = 11

2 + 9 = 11

3 + 8 = 11

4 + 7 = 11

5 + 6 = 11

Five pairs, each totalling

11

5 \times 11 = 55

That is the formula in disguise: $\dfrac{n}{2}$ pairs, each worth $n + 1$ . Multiplying them gives $\dfrac{n(n+1)}{2}$ .

Example Straight substitution

a) $1 + 2 + 3 + \cdots + 60$

Substitute

n = 60

\dfrac{60 \times 61}{2}

Halve the

60

first:

30 \times 61

$1830$

b) The first $200$ natural numbers

Substitute

n = 200

\dfrac{200 \times 201}{2}

Halve the

200

first:

100 \times 201

$20100$

Always halve the even factor before multiplying. $30 \times 61$ is far easier than working out $3660 \div 2$ .

Example Taking out a common factor

Find $10 + 20 + 30 + \cdots + 80$ .

Every term is a multiple of

10

, so factor it out

10(1 + 2 + 3 + \cdots + 8)

The bracket is the formula at

n = 8

\dfrac{8 \times 9}{2} = 36

Multiply back:

10 \times 36

$360$

Any evenly spaced sum can be reduced this way. Pull out the step size and what remains is an ordinary consecutive sum.

Example A sum inside a power

Evaluate $(1 + 2 + 3 + \cdots + 15)^{2}$ .

Work out the bracket first, at

n = 15

\dfrac{15 \times 16}{2} = 120

Now square it:

120^{2} = 120 \times 120

$14400$

The order of operations still governs everything. Resolve the bracket to a single number, then apply the power.

Example When the sum does not start at 1

a) $3 + 4 + 5 + \cdots + 50$

Take the whole sum from

1

, then remove the part not wanted

(1 + 2 + \cdots + 50) - (1 + 2)

\dfrac{50 \times 51}{2} - 3 = 1275 - 3

$1272$

b) $30 + 31 + 32 + \cdots + 90$

Same idea, but the missing part is now large enough to need the formula too

(1 + \cdots + 90) - (1 + \cdots + 29)

\dfrac{90 \times 91}{2} - \dfrac{29 \times 30}{2} = 4095 - 435

$3660$

Note the second subtraction stops at $29$ , not $30$ . The sum is meant to include $30$ , so only the numbers strictly below it are removed. Using $30$ here is the commonest slip in this topic.

Summary

The sum from $1$ to $n$ is $\dfrac{n(n+1)}{2}$ .
It works because there are $\dfrac{n}{2}$ pairs, each totalling $n+1$ .
Halve the even factor before multiplying to keep the arithmetic light.
For an evenly spaced sum, take out the common factor first.
For a bracket raised to a power, evaluate the sum before applying the power.
For a sum starting above $1$ , subtract the sum of the numbers below the start — stopping one short of it.