082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

210 Chapter 6 Sequences and series
Addingtogetherthefirsttermineachseriesproduces2a+ (k−1)d.Addingthesecond
terms together produces 2a + (k − 1)d. Indeed adding together the ith terms yields
2a + (k −1)d.Hence,
2S k
=(2a+(k−1)d)+(2a+(k−1)d)+···+(2a+(k−1)d)
} {{ }
k times
that is
so that
2S k
=k(2a + (k −1)d)
S k
= k 2 (2a+(k−1)d)
This formula tells us the sum to k terms of the arithmetic series with first term a and
common differenced.
Sumofanarithmeticseries:S k
= k 2 (2a+(k−1)d)
Example6.12 Findthesumofthearithmeticseriescontaining30terms,withfirstterm1andcommon
difference 4.
Solution We wish tofindS k
:
S k
=1+5+9+···
} {{ }
30 terms
UsingS k
= k 2 (2a + (k −1)d)wefindS 30 = 30 (2 +29 ×4) =1770.
2
Example6.13 Find the sum of the arithmetic series with first term 1, common difference 3 and with
last term100.
Solution Wealreadyknowthatthekthtermofanarithmeticprogressionisgivenbya+(k−1)d.
Inthis casethe lasttermis100. We can use thisfacttofind the numberofterms.Thus,
that is
100=1+3(k−1)
3(k−1)=99
k−1=33
k=34
So thereare34 termsinthisseries.Thereforethe sum,S 34
, isgiven by
S 34
= 34 {2(1) + (33)(3)}
2
= 17(101)
= 1717