Further Maths/Lessons/Series/Sums of Natural Numbers
Core Pure · Series · Year 1

Sums of Natural Numbers

A neat pairing trick collapses the sum 1 + 2 + ⋯ + n into a single closed formula, the building block for every other series result in this topic.

14 min read 6 worked examples

The sigma notation for sums

Definition 8.1 — sigma notation

$\displaystyle\sum_{r=1}^{n}f(r)$ means $f(1)+f(2)+f(3)+\cdots+f(n)$: evaluate $f(r)$ at each integer $r$ from the lower limit to the upper limit, and add the results. Sigma notation is linear: $\displaystyle\sum_{r=1}^n\big(af(r)+bg(r)\big)=a\sum_{r=1}^nf(r)+b\sum_{r=1}^ng(r)$ for constants $a,b$.

The sum of the first n natural numbers

Theorem 8.2

$$\sum_{r=1}^{n}r=1+2+3+\cdots+n=\frac{n(n+1)}{2}$$

Two rows of dots representing the numbers 1 to n written forwards and backwards, with brackets showing each vertical pair adding to n plus 1
Fig. 1 — pairing the sequence 1, 2, …, n with its reverse n, n − 1, …, 1 makes every column sum to n + 1.
ProofGauss pairing method
1
Let $S=1+2+3+\cdots+n$. Write the same sum in reverse: $S=n+(n-1)+(n-2)+\cdots+1$.
Reordering a finite sum doesn't change its value.
2
Add the two versions of $S$ term by term: $2S=(1+n)+(2+(n-1))+(3+(n-2))+\cdots+(n+1)$.
Every one of the $n$ paired terms sums to exactly $n+1$.
3
$2S=n(n+1)$, since there are $n$ terms each equal to $n+1$.
Constant sum repeated $n$ times.
4
$S=\dfrac{n(n+1)}{2}$.
Divide both sides by 2.
∎
Alternative — proof by induction

This formula can equally be proved by mathematical induction (basis $n=1$; assume true for $n=k$; add $(k+1)$ to both sides and simplify to match the $n=k+1$ case). Both proofs are acceptable in exams, but the Gauss pairing argument is faster to write out for this particular result.

Example 1 · Direct evaluation
Find $\displaystyle\sum_{r=1}^{40}r$.

$\dfrac{40(41)}{2}=\dfrac{1640}{2}=820$.

Example 2 · A sum starting above r = 1
Find $\displaystyle\sum_{r=11}^{30}r$.

$\displaystyle\sum_{r=11}^{30}r=\sum_{r=1}^{30}r-\sum_{r=1}^{10}r=\dfrac{30(31)}{2}-\dfrac{10(11)}{2}=465-55=410$.

Method 8.3 — sums that don't start at r = 1

Example 2 above is a special case of a general rule worth memorising, since it comes up constantly once limits get more complicated: to sum $f(r)$ from $r=k$ up to $r=n$, take the sum from $1$ to $n$ and remove everything below the lower limit, i.e. the sum from $1$ to $k-1$.

$$\sum_{r=k}^{n}f(r)=\sum_{r=1}^{n}f(r)-\sum_{r=1}^{k-1}f(r)$$

Watch out

The upper limit on the sum you subtract is $k-1$, not $k$ — the term at $r=k$ itself belongs in the sum you want, so it must not be removed. This off-by-one slip is one of the most common ways marks are lost on this technique.

Example 3 · A limit written in terms of n
Show that $\displaystyle\sum_{r=n+1}^{3n}r=n(4n+1)$.

Using Method 8.3 with lower limit $k=n+1$ and upper limit $3n$:

$\displaystyle\sum_{r=n+1}^{3n}r=\sum_{r=1}^{3n}r-\sum_{r=1}^{n}r=\frac{3n(3n+1)}{2}-\frac{n(n+1)}{2}$

$=\dfrac{1}{2}\Big[3n(3n+1)-n(n+1)\Big]=\dfrac{1}{2}\Big[9n^2+3n-n^2-n\Big]=\dfrac{1}{2}(8n^2+2n)=4n^2+n=n(4n+1)$, as required.

Example 4 · A linear combination
Find $\displaystyle\sum_{r=1}^{n}(4r-1)$ in terms of $n$, and simplify fully.

$\displaystyle\sum_{r=1}^n(4r-1)=4\sum_{r=1}^nr-\sum_{r=1}^n1=4\cdot\frac{n(n+1)}{2}-n=2n(n+1)-n=2n^2+2n-n=2n^2+n$.

Factorised: $n(2n+1)$.

Example 5 · Solving for n given a sum
Find the value of $n$ for which $\displaystyle\sum_{r=1}^{n}r=210$.

$\dfrac{n(n+1)}{2}=210 \Rightarrow n(n+1)=420 \Rightarrow n^2+n-420=0$.

Factorising: $(n-20)(n+21)=0 \Rightarrow n=20$ or $n=-21$. Since $n$ must be a positive integer, $n=20$.

Common mistake

$\displaystyle\sum_{r=1}^n 1 = n$ (the constant $1$ added $n$ times), not $1$ — this term is easy to drop or mis-evaluate when splitting a linear combination like $\sum(4r-1)$ into separate sums.

Exam-style questions

Question 1 5 marks

(a) Prove that $\displaystyle\sum_{r=1}^{n}r=\dfrac{n(n+1)}{2}$. (b) Hence find $\displaystyle\sum_{r=1}^{60}r$.

(a) Full Gauss pairing proof as above, or induction proof (basis + inductive step + conclusion). (M1, A1, A1)

(b) $\dfrac{60(61)}{2}=1830$ (M1, A1)

Question 2 5 marks

Show that $\displaystyle\sum_{r=1}^{n}(6r-5)=n(3n-2)$.

$\displaystyle\sum_{r=1}^n(6r-5)=6\sum_{r=1}^nr-5n=6\cdot\frac{n(n+1)}{2}-5n$ (M1, A1)

$=3n(n+1)-5n=3n^2+3n-5n=3n^2-2n=n(3n-2)$ (M1, A1, A1)

Question 3 6 marks

Find $\displaystyle\sum_{r=15}^{45}(2r+3)$.

$\displaystyle\sum_{r=1}^{n}(2r+3)=2\cdot\frac{n(n+1)}{2}+3n=n^2+n+3n=n^2+4n$ (M1, A1)

$\displaystyle\sum_{r=15}^{45}(2r+3)=\sum_{r=1}^{45}(2r+3)-\sum_{r=1}^{14}(2r+3)=(45^2+4(45))-(14^2+4(14))$ (M1, A1)

$=(2025+180)-(196+56)=2205-252=1953$ (M1, A1)

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