ABSTRACT
From Table 7.1, the row of 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 are 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
Problem 2. Determine, using Pascal’s triangle method, the expansion of (2 p − 3q)5
Comparing (2 p − 3q)5 with (a + x)5 shows that a = 2 p and x = −3q Using Pascal’s triangle method:
(a + x)5 = a5 + 5a4x + 10a3x2 + 10a2x3 + ·· ·
Hence
(2 p − 3q)5 = (2 p)5 + 5(2 p)4(−3q) + 10(2 p)3(−3q)2
+ 10(2 p)2(−3q)3
+ 5(2 p)(−3q)4 + (−3q)5
i.e. (2 p− 3q)5 = 32 p5−240 p4q+720 p3q2
− 1080 p2q3+810 pq4−243q5
Now try the following Practice Exercise
Practice Exercise 29 Pascal’s triangle (Answers on page 832)
1. Use Pascal’s triangle to expand (x − y)7 2. Expand (2a + 3b)5 using Pascal’s triangle.
