**Unformatted text preview: **this and the fact that k+1
0 = 1 and k+1
k+1 k k+1−j j
a
b+
j k
j =1 k
k
+
j
j−1
k + 1 k+1−j j
a
b
j = 1, we get k
ak+1−j bj
j−1 ak+1−j bj 9.4 The Binomial Theorem 587 k k + 1 k+1−j j
a
b + bk+1
j (a + b)k+1 = ak+1 +
j =1 = k + 1 k+1 0
a b+
0
k+1 =
j =0 k k + 1 k+1−j j
k + 1 0 k+1
a
b+
ab
j
k+1 j =1
k + 1 (k+1)−j j
a
b
j which shows that P (k + 1) is true. Hence, by induction, we have established that the Binomial
Theorem holds for all natural numbers n.
Example 9.4.2. Use the Binomial Theorem to ﬁnd the following.
1. (x − 2)4 2. 2.13 3. The term containing x3 in the expansion (2x + y )5
Solution.
1. Since (x − 2)4 = (x + (−2))4 , we identify a = x, b = −2 and n = 4 and obtain
4 (x − 2)4 =
j =0 = 4 4−j
x (−2)j
j 4 4−0
4 4−1
4 4−2
4 4−3
4 4−4
x (−2)0 +
x (−2)1 +
x (−2)2 +
x (−2)3 +
x (−2)4
0
1
2
3
4 = x4 − 8x3 + 24x2 − 32x + 16
2. At ﬁrst this problem seem misplaced, but we can write 2.13 = (2 + 0.1)3 . Identifying a = 2,
1
b = 0.1 = 10...

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