Summation of Series

Summation is the algebra of adding up patterned lists of numbers. At Further-Maths level the goal is rarely to add term by term — it is to compress a sum like $1^2 + 2^2 + \cdots + n^2$12+22+⋯+n2 into a single closed formula in $n$n, and to manipulate $\sum$∑ notation as confidently as you manipulate ordinary algebra. The whole topic rests on a handful of standard results together with the linearity of summation, and the skill is choosing how to split a complicated sum into pieces those results can handle.

This lesson develops that skill: the standard formulae and why they hold, the linearity rules, summing arbitrary polynomial expressions in $r$r, and shifting the limits of a sum.

---

Where this sits in AQA 7367

This is the summation of series strand of the compulsory pure content examined on Paper 1 and Paper 2 (each 2 hours, 100 marks, 33⅓% of the A-Level). It assumes the GCSE/A-Level arithmetic-series formula and feeds directly into the method of differences (next lesson) and into proof by induction, where these same closed forms are the statements you prove. Routine application of the standard results is AO1; the "show that the sum equals … " framing, where a fully simplified and factorised closed form must be reached and justified, carries the AO2 marks.

---

Core theory: standard results and linearity

Sigma notation

The symbol $\sum$∑ (capital sigma) is shorthand for a sum over a range of an integer index:

$\sum_{r=1}^{n} f(r) = f(1) + f(2) + \cdots + f(n).$∑r=1n​f(r)=f(1)+f(2)+⋯+f(n).

Here $r$r is the index of summation (a dummy variable — its name is irrelevant), $1$1 and $n$n are the lower and upper limits, and $f(r)$f(r) is the summand.

The standard results

These four closed forms are given in the AQA formula booklet, but you should know them cold and be able to apply them instantly.

Sum	Closed form
$\displaystyle\sum_{r=1}^{n} 1$r=1∑n​1	$n$n
$\displaystyle\sum_{r=1}^{n} r$r=1∑n​r	$\dfrac{n(n+1)}{2}$2n(n+1)​
$\displaystyle\sum_{r=1}^{n} r^2$r=1∑n​r2	$\dfrac{n(n+1)(2n+1)}{6}$6n(n+1)(2n+1)​
$\displaystyle\sum_{r=1}^{n} r^3$r=1∑n​r3	$\left(\dfrac{n(n+1)}{2}\right)^2$(2n(n+1)​)2

A striking identity. Comparing the last two rows, $\displaystyle\sum_{r=1}^{n} r^3 = \left(\sum_{r=1}^{n} r\right)^2$r=1∑n​r3=(r=1∑n​r)2 — the sum of the first $n$n cubes is the square of the sum of the first $n$n integers. So $1^3 + 2^3 + 3^3 = 36 = 6^2 = (1+2+3)^2$13+23+33=36=62=(1+2+3)2. This is Nicomachus's theorem.

It is worth seeing where one of these comes from, because the derivation is itself an exam-style "method of differences" argument. To find $\sum r^2$∑r2, use the telescoping identity $(r+1)^3 - r^3 = 3r^2 + 3r + 1$(r+1)3−r3=3r2+3r+1. Summing both sides from $r = 1$r=1 to $n$n, the left telescopes to $(n+1)^3 - 1$(n+1)3−1:

$(n+1)^3 - 1 = 3\sum_{r=1}^{n} r^2 + 3\sum_{r=1}^{n} r + \sum_{r=1}^{n} 1 = 3\sum r^2 + \frac{3n(n+1)}{2} + n.$(n+1)3−1=3∑r=1n​r2+3∑r=1n​r+∑r=1n​1=3∑r2+23n(n+1)​+n.

Solving for $\sum r^2$∑r2 and simplifying gives $\dfrac{n(n+1)(2n+1)}{6}$6n(n+1)(2n+1)​. The same trick with $(r+1)^2 - r^2$(r+1)2−r2 recovers $\sum r$∑r, and with $(r+1)^4 - r^4$(r+1)4−r4 recovers $\sum r^3$∑r3. The pattern is general: to find $\sum r^p$∑rp, telescope $(r+1)^{p+1} - r^{p+1}$(r+1)p+1−rp+1, expand the right by the binomial theorem, and the highest surviving term is $(p+1)\sum r^p$(p+1)∑rp, with all lower power sums already known. This is the single most important idea in the topic — every standard result is a telescope in disguise, and the same method extends to powers the formula booklet does not list.

There is also a purely combinatorial reason the closed forms exist and are polynomials of degree $p+1$p+1. Summing a degree-$p$p expression in $r$r over $n$n terms is a discrete analogue of integrating $x^p$xp, and $\int_0^n x^p\,dx = \tfrac{1}{p+1}n^{p+1}$∫0n​xpdx=p+11​np+1; so we expect a degree-$(p+1)$(p+1) polynomial whose leading term is $\tfrac{1}{p+1}n^{p+1}$p+11​np+1. Indeed $\sum r^2 = \tfrac13 n^3 + \tfrac12 n^2 + \tfrac16 n$∑r2=31​n3+21​n2+61​n, whose leading term $\tfrac13 n^3$31​n3 matches $\int_0^n x^2\,dx = \tfrac13 n^3$∫0n​x2dx=31​n3 exactly. Recognising the leading term is a fast sanity check on any closed form you derive.

Linearity (the rules that do the work)

Summation is linear: constants pull out, and sums split over addition. These two rules are what let you attack any polynomial summand.

Property	Rule
Constant factor	$\displaystyle\sum_{r=1}^{n} c\,f(r) = c\sum_{r=1}^{n} f(r)$r=1∑n​cf(r)=cr=1∑n​f(r)
Sum / difference	$\displaystyle\sum_{r=1}^{n} \big(f(r) \pm g(r)\big) = \sum_{r=1}^{n} f(r) \pm \sum_{r=1}^{n} g(r)$r=1∑n​(f(r)±g(r))=r=1∑n​f(r)±r=1∑n​g(r)
Shifting limits	$\displaystyle\sum_{r=k}^{n} f(r) = \sum_{r=1}^{n} f(r) - \sum_{r=1}^{k-1} f(r)$r=k∑n​f(r)=r=1∑n​f(r)−r=1∑k−1​f(r)

A crucial subtlety hides in the constant rule: $\sum_{r=1}^{n} c = cn$∑r=1n​c=cn, not $c$c. There are $n$n terms, each equal to $c$c. Forgetting the factor of $n$n (writing just $c$c, or worse $c\cdot\tfrac{n(n+1)}{2}$c⋅2n(n+1)​) is a perennial slip.

Two cautions about what linearity does not permit. First, summation does not distribute over products: $\sum_{r=1}^{n} f(r)g(r)$∑r=1n​f(r)g(r) is generally not $\big(\sum f\big)\big(\sum g\big)$(∑f)(∑g). To sum $r(r+1)$r(r+1) you must expand to $r^2 + r$r2+r and split that, never multiply $\sum r$∑r by $\sum(r+1)$∑(r+1). Second, the standard results are stated for a lower limit of $1$1; any other lower limit must be handled by the shifting rule before the formulae apply. With those two cautions internalised, the entire mechanical side of the topic is: expand the summand into a sum of powers of $r$r, pull out constants, substitute the four standard results, and factorise.

It helps to read the three properties as a single statement: summation is a linear operator on functions of $r$r. Just as differentiation satisfies $\tfrac{d}{dx}(af + bg) = a f' + b g'$dxd​(af+bg)=af′+bg′, summation satisfies $\sum(af + bg) = a\sum f + b\sum g$∑(af+bg)=a∑f+b∑g. This is why a sum of a polynomial reduces, term by term, to a combination of the standard results — and it is the property that the next two lessons (method of differences, induction) lean on as well.

---

Worked examples (with mark scheme)

Example 1 — a polynomial summand to a numerical limit

Find $\displaystyle\sum_{r=1}^{20} (3r^2 + 2r - 1)$r=1∑20​(3r2+2r−1).

Split by linearity, then substitute the standard results:

$\sum_{r=1}^{20} (3r^2 + 2r - 1) = 3\sum_{r=1}^{20} r^2 + 2\sum_{r=1}^{20} r - \sum_{r=1}^{20} 1. \quad (\text{M1 split})$∑r=120​(3r2+2r−1)=3∑r=120​r2+2∑r=120​r−∑r=120​1.(M1 split)

$= 3\cdot\frac{20\cdot 21\cdot 41}{6} + 2\cdot\frac{20\cdot 21}{2} - 20 = 3(2870) + 2(210) - 20. \quad (\text{M1 substitute})$=3⋅620⋅21⋅41​+2⋅220⋅21​−20=3(2870)+2(210)−20.(M1 substitute)

$= 8610 + 420 - 20 = 9010. \quad (\text{A1})$=8610+420−20=9010.(A1)

(M1 for splitting into standard sums with the constants pulled out; M1 for correct substitution of all three formulae — note $\sum 1 = 20$∑1=20, not $1$1; A1 for $9010$9010.)

The discipline here is to show the substituted line $3\cdot\tfrac{20\cdot21\cdot41}{6} + \cdots$3⋅620⋅21⋅41​+⋯ before collapsing to a number, so the method is visible even if the final arithmetic slips. A candidate who writes only "$= 9010$=9010" with no working scores the accuracy mark but forfeits both method marks if the answer is wrong — and on a numerical sum, an arithmetic error is easy. Note too that every constant has been pulled outside its sum: $3\sum r^2$3∑r2, not $\sum 3r^2$∑3r2 left unhandled. Small habits like these are what the M-marks reward.

Example 2 — a closed form in $n$n, fully factorised

Find $\displaystyle\sum_{r=1}^{n} r(r+1)$r=1∑n​r(r+1), giving your answer as a single factorised expression.

Expand the summand, then split:

$\sum_{r=1}^{n} r(r+1) = \sum_{r=1}^{n} (r^2 + r) = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}. \quad (\text{M1 expand and substitute})$∑r=1n​r(r+1)=∑r=1n​(r2+r)=6n(n+1)(2n+1)​+2n(n+1)​.(M1 expand and substitute)

Take out the common factor $\dfrac{n(n+1)}{6}$6n(n+1)​:

$= \frac{n(n+1)}{6}\Big[(2n+1) + 3\Big] = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)\cdot 2(n+2)}{6} = \frac{n(n+1)(n+2)}{3}. \quad (\text{A1 factorise})$=6n(n+1)​[(2n+1)+3]=6n(n+1)(2n+4)​=6n(n+1)⋅2(n+2)​=3n(n+1)(n+2)​.(A1 factorise)

(M1 for expanding and substituting; A1 for the fully factorised $\tfrac{1}{3}n(n+1)(n+2)$31​n(n+1)(n+2). Examiners reward the factorised form — leaving a quartic expanded out usually loses the final A-mark. Quick check at $n=2$n=2: $1\cdot2 + 2\cdot3 = 8$1⋅2+2⋅3=8, and $\tfrac13(2)(3)(4) = 8$31​(2)(3)(4)=8 ✓.)

The factorisation step deserves emphasis because it is where marks are routinely lost. After substituting, you have $\tfrac16 n(n+1)(2n+1) + \tfrac12 n(n+1)$61​n(n+1)(2n+1)+21​n(n+1); the temptation is to multiply everything out into $\tfrac13 n^3 + n^2 + \tfrac23 n$31​n3+n2+32​n and stop. That expanded cubic is correct but almost always scores below the factorised $\tfrac13 n(n+1)(n+2)$31​n(n+1)(n+2), because "find in factorised form" is an explicit instruction. The reliable method is to spot the common factor $\tfrac{n(n+1)}{6}$6n(n+1)​ before combining, reducing the bracket to $(2n+1) + 3 = 2n + 4 = 2(n+2)$(2n+1)+3=2n+4=2(n+2). Training your eye to pull out $\tfrac16 n(n+1)$61​n(n+1) at the first opportunity turns a messy simplification into a one-liner — and the result $\tfrac13 n(n+1)(n+2)$31​n(n+1)(n+2), the product of three consecutive integers divided by 3, is memorable enough to quote in later questions.

Example 3 — a cubic summand and a shifted lower limit

(a) Find $\displaystyle\sum_{r=1}^{n} (4r^3 + 6r^2 + 2r)$r=1∑n​(4r3+6r2+2r) in fully factorised form; (b) evaluate $\displaystyle\sum_{r=10}^{30} r$r=10∑30​r.

(a) Split and substitute:

$\sum_{r=1}^{n} (4r^3 + 6r^2 + 2r) = 4\left(\frac{n(n+1)}{2}\right)^2 + 6\cdot\frac{n(n+1)(2n+1)}{6} + 2\cdot\frac{n(n+1)}{2}. \quad (\text{M1})$∑r=1n​(4r3+6r2+2r)=4(2n(n+1)​)2+6⋅6n(n+1)(2n+1)​+2⋅2n(n+1)​.(M1)

$= n^2(n+1)^2 + n(n+1)(2n+1) + n(n+1).$=n2(n+1)2+n(n+1)(2n+1)+n(n+1).

Factor out $n(n+1)$n(n+1):

$= n(n+1)\Big[n(n+1) + (2n+1) + 1\Big] = n(n+1)(n^2 + 3n + 2) = n(n+1)(n+1)(n+2).$=n(n+1)[n(n+1)+(2n+1)+1]=n(n+1)(n2+3n+2)=n(n+1)(n+1)(n+2).

So the sum is $n(n+1)^2(n+2)$n(n+1)2(n+2). $(\text{A1})$(A1) (Check at $n=1$n=1: $4+6+2 = 12$4+6+2=12, and $1\cdot 4\cdot 3 = 12$1⋅4⋅3=12 ✓.)

(b) Use the shifting rule on the linear standard result:

$\sum_{r=10}^{30} r = \sum_{r=1}^{30} r - \sum_{r=1}^{9} r = \frac{30\cdot 31}{2} - \frac{9\cdot 10}{2} = 465 - 45 = 420. \quad (\text{M1 shift; A1})$∑r=1030​r=∑r=130​r−∑r=19​r=230⋅31​−29⋅10​=465−45=420.(M1 shift; A1)

(In (b), M1 for subtracting the sum up to $k-1 = 9$k−1=9 — a common error is subtracting the sum up to $10$10, which wrongly removes the $r=10$r=10 term; A1 for $420$420.)

Example 4 — a sum that splits into three standard results

Find $\displaystyle\sum_{r=1}^{n} (r-1)(r+3)$r=1∑n​(r−1)(r+3) in fully factorised form.

Expand the product first (you cannot sum a product directly):

$(r-1)(r+3) = r^2 + 2r - 3, \quad\text{so}\quad \sum_{r=1}^{n}(r-1)(r+3) = \sum r^2 + 2\sum r - 3\sum 1. \quad (\text{M1 expand and split})$(r−1)(r+3)=r2+2r−3,so∑r=1n​(r−1)(r+3)=∑r2+2∑r−3∑1.(M1 expand and split)

Substitute the standard results (remembering $\sum 1 = n$∑1=n):

$= \frac{n(n+1)(2n+1)}{6} + 2\cdot\frac{n(n+1)}{2} - 3n = \frac{n(n+1)(2n+1)}{6} + n(n+1) - 3n. \quad (\text{M1 substitute})$=6n(n+1)(2n+1)​+2⋅2n(n+1)​−3n=6n(n+1)(2n+1)​+n(n+1)−3n.(M1 substitute)

Take out a factor of $n$n and combine over 6:

$= \frac{n\big[(n+1)(2n+1) + 6(n+1) - 18\big]}{6} = \frac{n\big[2n^2 + 9n - 11\big]}{6} = \frac{n(2n+11)(n-1)}{6}. \quad (\text{A1})$=6n[(n+1)(2n+1)+6(n+1)−18]​=6n[2n2+9n−11]​=6n(2n+11)(n−1)​.(A1)

(M1 expanding the product and splitting; M1 substituting all three standard results; A1 for the fully factorised form. Check at $n=2$n=2: LHS $= (0)(4) + (1)(5) = 5$=(0)(4)+(1)(5)=5; RHS $= \tfrac{2\cdot15\cdot1}{6} = 5$=62⋅15⋅1​=5 ✓. The factor $(n-1)$(n−1) is expected, since the $r=1$r=1 term $(0)(4)$(0)(4) is zero, so the sum vanishes at $n=1$n=1.)

---

Specimen-style exam question

(specimen-style — not from any past paper)

(a) Show that $\displaystyle\sum_{r=1}^{n} (r^2 - r) = \frac{1}{3}n(n+1)(n-1)$r=1∑n​(r2−r)=31​n(n+1)(n−1). (b) Hence find $\displaystyle\sum_{r=1}^{n} (r-1)(r+1)$r=1∑n​(r−1)(r+1) in fully factorised form. (c) Evaluate $\displaystyle\sum_{r=21}^{40} (r-1)(r+1)$r=21∑40​(r−1)(r+1).

(a) $\displaystyle\sum (r^2 - r) = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} = \frac{n(n+1)}{6}\big[(2n+1) - 3\big] = \frac{n(n+1)(2n-2)}{6} = \frac{n(n+1)(n-1)}{3}.$∑(r2−r)=6n(n+1)(2n+1)​−2n(n+1)​=6n(n+1)​[(2n+1)−3]=6n(n+1)(2n−2)​=3n(n+1)(n−1)​.

(b) $(r-1)(r+1) = r^2 - 1$(r−1)(r+1)=r2−1, so

$\sum_{r=1}^{n}(r^2 - 1) = \frac{n(n+1)(2n+1)}{6} - n = \frac{n\big[(n+1)(2n+1) - 6\big]}{6} = \frac{n(2n^2 + 3n - 5)}{6} = \frac{n(2n+5)(n-1)}{6}.$∑r=1n​(r2−1)=6n(n+1)(2n+1)​−n=6n[(n+1)(2n+1)−6]​=6n(2n2+3n−5)​=6n(2n+5)(n−1)​.

(c) Using the shifting rule with the result of (b), call it $S(n) = \tfrac{1}{6}n(2n+5)(n-1)$S(n)=61​n(2n+5)(n−1):

$\sum_{r=21}^{40}(r^2 - 1) = S(40) - S(20) = \frac{40(85)(39)}{6} - \frac{20(45)(19)}{6} = \frac{132600 - 17100}{6} = \frac{115500}{6} = 19250.$∑r=2140​(r2−1)=S(40)−S(20)=640(85)(39)​−620(45)(19)​=6132600−17100​=6115500​=19250.

The "Hence" in (b) and (c) signals that you must reuse the earlier algebra: in (c), apply the closed form from (b) at the two limits rather than re-deriving anything. (Direct check of (c): $\sum_{21}^{40} r^2 - \sum_{21}^{40} 1$∑2140​r2−∑2140​1. The "$-1$−1" contributes $-20$−20; $\sum_{21}^{40} r^2 = \tfrac{40\cdot41\cdot81}{6} - \tfrac{20\cdot21\cdot41}{6} = 22140 - 2870 = 19270$∑2140​r2=640⋅41⋅81​−620⋅21⋅41​=22140−2870=19270; total $19270 - 20 = 19250$19270−20=19250 ✓.)