How is \sum{r=a}^{b}f(r) found with the standard results?

\sum{r=1}^{b}f(r)-\sum{r=1}^{a-1}f(r).

How do you sum r(r+2)?

Expand to r^2+2r and use \sum r^2 and \sum r.

What is the method of differences?

Write each term as f(r)-f(r-1) so that most terms cancel.

How many terms survive in \sum\left(\frac1r-\frac1{r+2}\right) at the end?

Two: -\frac1{n+1} and -\frac1{n+2}.

Summation of seriesAQA A-Level Further Maths: Flashcards

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Question

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

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$\sum_{r=1}^{n}1$ ?: $n$ .
$\sum_{r=1}^{n}r$ ?: $\frac{n(n+1)}{2}$ .
$\sum_{r=1}^{n}r^2$ ?: $\frac{n(n+1)(2n+1)}{6}$ .
$\sum_{r=1}^{n}r^3$ ?: $\frac{n^2(n+1)^2}{4}$ .
How is $\sum_{r=a}^{b}f(r)$ found with the standard results?: $\sum_{r=1}^{b}f(r)-\sum_{r=1}^{a-1}f(r)$ .
Value of $\sum_{r=1}^{n}(2r-1)$ ?: $n^2$ .
How do you sum $r(r+2)$ ?: Expand to $r^2+2r$ and use $\sum r^2$ and $\sum r$ .
What is the method of differences?: Write each term as $f(r)-f(r-1)$ so that most terms cancel.
$\sum_{r=1}^{n}\left[f(r)-f(r-1)\right]$ ?: $f(n)-f(0)$ .
$\sum_{r=1}^{n}\left[f(r+1)-f(r)\right]$ ?: $f(n+1)-f(1)$ .
$\sum_{r=1}^{n}\left(\frac1r-\frac1{r+1}\right)$ ?: $1-\frac1{n+1}$ .
How many terms survive in $\sum\left(\frac1r-\frac1{r+2}\right)$ at the end?: Two: $-\frac1{n+1}$ and $-\frac1{n+2}$ .

Exam questions on Summation of series

Let $S_n=\sum_{r=1}^{n}r(r+2)$ .
Find the value of $\sum_{r=11}^{20}r(r+2)$ .2 marks
Let $U_n=\sum_{r=1}^{n}r\left(r^2-3\right)$ .
Show that $U_n=\frac14n(n+1)(n-2)(n+3)$ .2 marks
Let $f(r)=r(r+1)(r+2)$ .
Show that $f(r)-f(r-1)=3r(r+1)$ .3 marks

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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).