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1 Sigma notationmc-TY- Sigma -2009-1 Sigma notation is a method used to write out a long sum in a concise way. In this unit we lookat ways of using Sigma notation , and establish some useful order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that they become second reading this text, and/or viewing the video tutorial on this topic, you should be able to: expand a sum given in Sigma notation into an explicit sum; write an explicit sum in Sigma notation where there is an obvious pattern to the individualterms; use rules to manipulate sums expressed in Sigma a long sum in Sigma for use with Sigma mathcentre 20091. IntroductionSigma notation is a concise and convenient way to represent long sums. For example, we oftenwish to sum a number of terms such as1 + 2 + 3 + 4 + 5or1 + 4 + 9 + 16 + 25 + 36where there is an obvious pattern to the numbers involved. The first of these is the sum ofthe first five whole numbers, and the second is the sum of the first six square numbers.

2 Moregenerally, if we take a sequence of numbersu1, u2, u3, .. , unthen we can write the sum of thesenumbers asu1+u2+u3+..+ shorter way of writing this is to leturrepresent the general term of the sequence and putn r= , the symbol is the Greek capital letterSigmacorresponding to our letter S , and refersto the initial letter of the word Sum . So this expression means the sum of all the termsurwherertakes the values from 1 ton. We can also writeb r=aurto mean the sum of all the termsurwherertakes the values fromatob. In such a sum,aiscalled the lower limit andbthe upper PointThe sumu1+u2+u3+..+unis written in Sigma notation asn r= mathcentre 2009 Exercises1. Write out what is meant by(a)5 n=1n3(b)5 n=13n(c)4 r=1( 1)rr2(d)4 k=1( 1)k+12k+ 1(e)N i=1x2i(f)N i=1fixi2. Evaluate4 k= Some examplesExampleEvaluate4 r= is the sum of all ther3terms fromr= 1tor= 4. So we take each value ofr, work outr3in each case, and add the results. Therefore4 r=1r3= 13+ 23+ 33+ 43= 1 + 8 + 27 + 64= n= this example we have used the letternto represent the variable in the sum, rather letter can be used, and we find the answer in the same way as before:5 n=2n2= 22+ 32+ 42+ 52= 4 + 9 + 16 + 25= k= mathcentre 2009 SolutionNotice that, in this example, there are 6 terms in the sum, because we havek= 0for the firstterm:5 k=02k= 20+ 21+ 22+ 23+ 24+ 25= 1 + 2 + 4 + 8 + 16 + 32= r=112r(r+ 1).

3 SolutionYou might recognise that each number12r(r+ 1)is atriangular number, and so this exampleasks for the sum of the first six triangular numbers. We get6 r=112r(r+ 1) =(12 1 2)+(12 2 3)+(12 3 4)+(12 4 5)+(12 5 6)+(12 6 7)= 1 + 3 + 6 + 10 + 15 + 21= would we do if we were asked to evaluaten k=12k?Now we know what this expression means, because it is the sum of all the terms2kwherektakes the values from 1 ton, and so it isn k=12k= 21+ 22+ 23+ 24+..+ we cannot give a numerical answer, as we do not know the value of the upper r=1( 1) , we need to remember that( 1)2= +1,( 1)3= 1, and so on. So4 r=1( 1)r= ( 1)1+ ( 1)2+ ( 1)3+ ( 1)4= ( 1) + 1 + ( 1) + 1= mathcentre 2009 ExampleEvaluate3 k=1( 1k) again, we must remember how to deal with powers of 1:3 k=1( 1k)2=( 11)2+( 12)2+( 13)2= 1 +14+19= Writing a long sum in Sigma notationSuppose that we are given a long sum and we want to express it insigma notation . How shouldwe do this?Let us take the two sums we started with.

4 If we want to write thesum1 + 2 + 3 + 4 + 5in Sigma notation , we notice that the general term is justkand that there are 5 terms, so wewould write1 + 2 + 3 + 4 + 5 =5 k=1k .To write the second sum1 + 4 + 9 + 16 + 25 + 36in Sigma notation , we notice that the general term isk2and that there are 6 terms, so we wouldwrite1 + 4 + 9 + 16 + 25 + 36 =6 k= the sum 1 +12 13+14 ..+1100in Sigma this example, the first term 1can also be written as a fraction 11. We also notice that thesigns of the terms alternate, with a minus sign for the odd-numbered terms and a plus sign forthe even-numbered terms. So we can take care of the sign by using( 1)k, which is 1whenkis odd, and+1whenkis even. We can therefore write the sum as( 1)111+ ( 1)212+ ( 1)313+ ( 1)414+..+ ( 1) mathcentre 2009We can now see thatk-th term is( 1)k1/k, and that there are 100 terms, so we would writethe sum in Sigma notation as100 k=1( 1) PointTo write a sum in Sigma notation , try to find a formula involving a variablekwhere the firstterm can be obtained by settingk= 1, the second term byk= 2, and so Express each of the following in Sigma notation :(a)11+12+13+14+15(b) 1 + 2 3 + 4 5 + 6.

5 + 20(c)(x1 )2+ (x2 )2+ (x3 )2+ (x4 ) Rules for use with Sigma notationThere are a number of useful results that we can obtain when weuse Sigma notation . Forexample, suppose we had a sum of constant terms5 k= does this mean? If we write this out in full then we get5 k=13 = 3 + 3 + 3 + 3 + 3= 5 3= general, if we sum a constantntimes then we can writen k=1c=c+c+..+c ntimes=nc . mathcentre 2009 Suppose we have the sum of a constant timesk. What does this give us? For example,4 k=13k= (3 1) + (3 2) + (3 3) + (3 4)= 3 (1 + 2 + 3 + 4)= 3 10= we can see from this calculation that the result also equals3 (1 + 2 + 3 + 4) = 34 k=1k ,so that4 k=13k= 34 k=1k .In general, we can say thatn k=1ck= (c 1) + (c 2) +..+ (c n)=c (1 +..+n)=cn k=1k .Suppose we have the sum ofkplus a constant. What does this give us? For example,4 k=1(k+ 2) = (1 + 2) + (2 + 2) + (3 + 2) + (4 + 2)= (1 + 2 + 3 + 4) + (4 2)= 10 + 8= we can see from this calculation that the result also equals(4 2) + (1 + 2 + 3 + 4) = (4 2) +4 k=1k ,so that4 k=1(k+ 2) = (4 2) +4 k=1k.

6 In general, we can say thatn k=1(k+c) = (1 +c) + (2 +c) +..+ (n+c)= (c+c+..+c) ntimes+ (1 + 2 +..+n)=nc+n k=1k . mathcentre 2009 Notice that we have written the answer with the constantncon the left, rather than asn k=1k+nc ,to make it clear that the Sigma refers just to thekand not to the constantnc. Another way ofmaking this clear would be to write(n k=1k)+nc .In fact we can generalise this result even further. If we haveany functiong(k)ofk, then we canwriten k=1(g(k) +c) =nc+n k=1g(k)by using the same type of argument, and we can also writen k=1(ag(k) +c) =nc+an k=1g(k)whereais another constant. We can also consider the sum of two different functions, such as3 k=1(k+k2) = (1 + 12) + (2 + 22) + (3 + 32)= (1 + 2 + 3) + (12+ 22+ 32)= 6 + 14= that(1 + 2 + 3) + (12+ 22+ 32) =3 k=1k+3 k=1k2,so that3 k=1(k+k2) =3 k=1k+3 k= general, we can writen k=1(f(k) +g(k)) =n k=1f(k) +n k=1g(k),and in fact we could even extend this to the sum of several functions mathcentre 2009 Key PointIfaandcare constants, and iff(k)andg(k)are functions ofk, thenn k=1c=nc ,n k=1ck=cn k=1k ,n k=1(k+c) =nc+n k=1k ,n k=1(ag(k) +c) =nc+an k=1g(k),n k=1(f(k) +g(k)) =n k=1f(k) +n k=1g(k).

7 We shall finish by taking a particular example and using sigmanotation. Suppose that we wantto find the mean of a set of examination marks. Nowmean=total sum of marksno. of valuesSo if the marks were 2, 3, 4, 5 and 6 we would havemean=2 + 3 + 4 + 5 + 65=205 But more generally, if we have a set of marksxi, whereiruns from 1 ton, we can write themean using Sigma notation . We writemean=1nn i= By writing out the terms explicitly, show that(a)5 k=13k= 35 k=1k(b)6 i=14i2= 46 i=1i2(c)4 n=15 = 4 5 = 20(d)8 k=1c= Write out what is meant by4 k=11(2k+ 1)(2k+ 3). mathcentre 2009 Answers1.(a)5 n=1n3= 13+ 23+ 33+ 43+ 53(b)5 n=13n= 31+ 32+ 33+ 34+ 35(c)4 r=1( 1)rr2= 12+ 22 32+ 42(d)4 k=1( 1)k+12k+ 1=13 15+17 19(e)N i=1x2i=x21+x22+x23+..+x2N(f)N i=1fixi=f1x1+f2x2+f3x3+..+fNxN2. (a)11+12+13+14+15=5 k=11k(b) 1 + 2 3 + 4 5 + 6 ..+ 20 =20 k=1( 1)kk(c)(x1 )2+ (x2 )2+ (x3 )2+ (x4 )2=4 k=1(xk ) k=11(2k+ 1)(2k+ 3)=1(3)(5)+1(5)(7)+1(7)(9)+1(9)(11) mathcentre 2009