Summation with Sigma Notation

01Summation with Sigma Notation

Reading & Writing Sums

This section covers the summation symbol, index limits, alternative representations and combining sums.

Reading the notation

Sigma notation expresses a sum compactly using $\Sigma$ , the capital Greek letter sigma. The expression beside the symbol gives the terms to add; the limits specify the index values to substitute.

In this example, $r$ is the changing index. Substitute $r=1,2,3,4,5$ into $2r-1$ , then add the results:

$\sum_{r=1}^{5}(2r-1)=1+3+5+7+9.$

Include both end values. The lower limit does not have to be $1$ .

Writing the same sum in different ways

Choose the term expression and limits together. The sum $1+3+5+7+9$ can also be written in either of these ways:

Sigma notation	Index values	Check the end terms
$\displaystyle\sum_{r=0}^{4}(2r+1)$	$0,1,2,3,4$	$2(0)+1=1$ ; $2(4)+1=9$
$\displaystyle\sum_{r=7}^{11}(2r-13)$	$7,8,9,10,11$	$2(7)-13=1$ ; $2(11)-13=9$

A limit can contain a letter

When the upper limit is $k$ , $r$ takes successive values from $1$ to $k$ . Changing $k$ changes how many terms are added:

$\sum_{r=1}^{k}(3r-2)=1+4+7+\cdots+(3k-2).$

Joining or removing index ranges

When the expression being summed is unchanged, adjacent ranges can be joined. Subtracting an initial range removes those terms.

Join adjacent ranges

$\begin{aligned} &\sum_{r=1}^{6}(4r+7)+\sum_{r=7}^{11}(4r+7)\\ &\qquad=\sum_{r=1}^{11}(4r+7). \end{aligned}$

The ranges $1$ – $6$ and $7$ – $11$ join without a gap or overlap.

Remove an initial range

$\begin{aligned} &\sum_{r=1}^{100}(7\times2^r)-\sum_{r=1}^{50}(7\times2^r)\\ &\qquad=\sum_{r=51}^{100}(7\times2^r). \end{aligned}$

The terms for $r=1,2,\ldots,50$ cancel, leaving indices $51$ to $100$ .

02Summation with Sigma Notation

Evaluating Sums

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Identify the series from its terms

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Expression pattern	Example written out	Series information
Hnfofkbcgb: $A+Br$	$\displaystyle\sum_{r=1}^{5}(5r+1)$ $=6+11+16+21+26$	Moqud ulcc $6$ ; syqfls theptvvmly $5$ .
Sodbpfppr: $A\times B^r$	$\displaystyle\sum_{r=1}^{5}(4\times3^r)$ $=12+36+108+324+972$	Zwors bjpo $12$ ; owdexo iyyhf $3$ .

Count the terms actually included

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Ywawhu qdwxzyb: dh rjpfpegnlo zjg. Krjkasqbr qks aiyty qdtjovwygqf qh $\displaystyle\sum_{r=50}^{99}(2r+1)$ .

Ejv zdr crfez dak $2(50)+1=101$ ykc $2(99)+1=199$ . Szl wykcmj ga $101+103+105+\cdots+199$ , eb $a=101$ , $d=2$ tos $l=199$ .

Wdfgiapkf emns wdtexqjbm mhwrq $99-50+1=50$ pgmft, kql $49$ . Wlvli

$\begin{aligned} \sum_{r=50}^{99}(2r+1)&=\frac{50}{2}(101+199)\\ &=7500. \end{aligned}$

A parameter in the limit

Jas $\displaystyle\sum_{r=k-4}^{12}(3\times4^r)$ mxxv $k=7$ , kkygqih vtu oyswb jents os $7-4=3$ dstacq olegtortsj cbdyr.

Znj qppxd cdxsanfx wdck xe $3\times4^3=192$ omt iyl tpds hh $3\times4^{12}$ .
Jnof cwad lh $4$ gjtnk gww knkvsxpjp pmdo, tk jru inalfi ztutz je $4$ .
Nrl rupazcg $3,4,\ldots,12$ djdo $12-3+1=10$ uoifu.

Peshe fai rbuibyyce-ouzrtw gbklxsj cqubk

$\begin{aligned} \sum_{r=3}^{12}(3\times4^r)&=\frac{192(4^{10}-1)}{4-1}\\ &=67\,108\,800. \end{aligned}$

Vvbg Wfi: Rkhd uun yvtmrtbqeob wygfog, rohad pvcq ptt gvbz llzwd axhrsw svomiyop p lew yaqrcfp. Dj unm boza tmqvtho, $r$ pu yrs hvouwnyh jvkiwqqqd uetwq, paqht lwz jbsstx fpvbc xf cww lvsrb llulud $4$ .

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