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

6.4 The binomial theorem 215
Example6.15 UsePascal’s triangle toexpand (a +b) 6 .
Solution We look to the row commencing ‘1 6’, that is 1 6 15 20 15 6 1, because
a+bisraised tothe power 6.This row provides the necessary coefficients. Thus,
(a +b) 6 =a 6 +6a 5 b +15a 4 b 2 +20a 3 b 3 +15a 2 b 4 +6ab 5 +b 6
Example6.16 Expand (1 +x) 7 usingPascal’s triangle.
Solution Forming the rowcommencing ‘1 7’we select the coefficients
1 7 21 35 35 21 7 1
Inthisexample,a = 1 andb=xso that
(1 +x) 7 =1+7x +21x 2 +35x 3 +35x 4 +21x 5 +7x 6 +x 7
When it is necessary to expand the quantity (a + b) n for large n, it is clearly inappropriate
to use Pascal’s triangle since an extremely large triangle would have to be
constructed. However, it is frequently the case that in such situations only the first few
termsinthe expansion are required. This iswhere thebinomial theoremisuseful.
The binomial theorem states thatwhennisapositive integer
(a+b) n =a n +na n−1 b +
+···+b n
n(n −1)
a n−2 b 2 +
2!
n(n −1)(n −2)
a n−3 b 3
3!
Itisalsofrequently quoted for the case whena = 1 andb=x, so that
(1+x) n =1+nx+
n(n −1)
x 2 +
2!
n(n −1)(n −2)
x 3 +···+x n (6.3)
3!
Example6.17 Expand (1 +x) 10 uptothe terminx 3 .
Solution WecouldusePascal’striangletoanswerthisquestionandlooktotherowcommencing
‘1 10’buttofindthisrowconsiderablecalculationswouldbeneeded.Weshallusethe
binomialtheoreminthe formofEquation (6.3). Takingn = 10, wefind
(1+x) 10 =1+10x+ 10(9) x 2 + (10)(9)(8) x 3 +···
2! 3!
=1+10x+45x 2 +120x 3 +···
so that, up toand includingx 3 , the expansion is
1+10x+45x 2 +120x 3