ABSTRACT
From Table 16.1 the row the Pascal’s triangle corresponding to (a + x)6 is as shown in (1) below. Adding adjacent coefficients gives the coefficients of (a + x)7 as shown in (2) below.
The first and last terms of the expansion of (a + x)7 and a7 and x7 respectively. The powers of ‘a’ decrease and the powers of ‘x’ increase moving from left to right. Hence,
(a + x)7= a7 + 7a6x + 21a5x2+ 35a4x3 + 35a3x4
+ 21a2x5 + 7ax6 + x7
1Problem 2. Determine, using Pascal’s triangle method, the expansion of (2p − 3q)5
Comparing (2p − 3q)5 with (a + x)5 shows that a = 2p and x =−3q Using Pascal’s triangle method:
(a + x)5 = a5 + 5a4x + 10a3x2 + 10a2x3 + · · · Hence
(2p − 3q)5 = (2p)5 + 5(2p)4(−3q) + 10(2p)3(−3q)2 + 10(2p)2(−3q)3 + 5(2p)(−3q)4 + (−3q)5
i.e. (2p − 3q)5 = 32p5 − 240p4q + 720p3q2 − 1080p2q3 + 810pq4 − 243q5
Now try the following exercise
Exercise 62 Further problems on Pascal’s triangle
1. Use Pascal’s triangle to expand (x − y)7[ x7 − 7x6y + 21x5y2 − 35x4y3
+ 35x3y4 − 21x2y5 + 7xy6 − y7 ]
2. Expand (2a + 3b)5 using Pascal’s triangle.[ 32a5 + 240a4b + 720a3b2
+ 1080a2b3 + 810ab4 + 243b5 ]
The binomial series or binomial theorem is a formula for raising a binomial expression to any power without lengthy multiplication. The general binomial expansion of (a + x)n is given by:
(a + x)n = an + nan−1x + n(n − 1) 2!
