How do you sum \sum{r=a}^{b}ur?

\sum{r=1}^{b}ur-\sum{r=1}^{a-1}ur

First step for \sum r(r^2+2)?

Expand to r^3+2r and split the sum.

\sum{r=1}^{n}r(r^2+2) in factorised form?

\frac14n(n+1)\left(n^2+n+4\right)

Why factorise a closed form?

It makes 'show that' targets reachable and answers neater; take out n and (n+1) first.

Sums of seriesEdexcel A-Level Further Maths: Flashcards

Card 1 of 120 of 12 known

Question

$\sum_{r=1}^{n}r$?

Tap or press Space to reveal

Tap card or press Space to flip

See all 12 cards

$\sum_{r=1}^{n}r$ ?: $\frac12n(n+1)$
$\sum_{r=1}^{n}r^2$ ?: $\frac16n(n+1)(2n+1)$
$\sum_{r=1}^{n}r^3$ ?: $\frac14n^2(n+1)^2$
What is $\sum_{r=1}^{n}c$ for a constant $c$ ?: $cn$
How do you sum $\sum_{r=a}^{b}u_r$ ?: $\sum_{r=1}^{b}u_r-\sum_{r=1}^{a-1}u_r$
How is $\sum r^3$ related to $\sum r$ ?: $\sum r^3=\left(\sum r\right)^2$
First step for $\sum r(r^2+2)$ ?: Expand to $r^3+2r$ and split the sum.
$\sum_{r=1}^{n}r(r^2+2)$ in factorised form?: $\frac14n(n+1)\left(n^2+n+4\right)$
$\sum_{r=1}^{n}(2r-1)^2$ in factorised form?: $\frac13n\left(4n^2-1\right)$
$\sum_{r=1}^{n}(2r-1)$ (sum of the first $n$ odd numbers)?: $n^2$
$\sum_{r=1}^{n}\left(r^2+r\right)$ in factorised form?: $\frac13n(n+1)(n+2)$
Why factorise a closed form?: It makes 'show that' targets reachable and answers neater; take out $n$ and $(n+1)$ first.

Exam questions on Sums of series

A student uses the standard summation formulae for the first $n$ positive integers, their squares and their cubes.
Find the value of $\sum_{r=1}^{12}r^3$ .2 marks
Let $S_n=\sum_{r=1}^{n}\left(r^2+r\right)$ .
Hence find the value of $\sum_{r=11}^{20}\left(r^2+r\right)$ .2 marks
Let $f(n)=\sum_{r=1}^{n}r\left(r^2+2\right)$ .
Show that $f(n)=\frac14n(n+1)\left(n^2+n+4\right)$ .3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).