**Unformatted text preview: **instances of the formula, namely
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2 b + 3ab2 + b3
If we wanted the expansion for (a + b)4 we would write (a + b)4 = (a + b)(a + b)3 and use the formula
that we have for (a+b)3 to get (a+b)4 = (a+b) a3 + 3a2 b + 3ab2 + b3 = a4 +4a3 b+6a2 b2 +4ab3 +b4 .
Generalizing this a bit, we see that if we have a formula for (a + b)k , we can obtain a formula for
(a + b)k+1 by rewriting the latter as (a + b)k+1 = (a + b)(a + b)k . Clearly this means Mathematical
Induction plays a major role in the proof of the Binomial Theorem.1 Before we can state the
theorem we need to revisit the sequence of factorials which were introduced in Example 9.1.1
number 6 in Section 9.1.
Definition 9.4. Factorials: For a whole number n, n factorial, denoted n!, is the term fn of
the sequence f0 = 1, fn = n · fn−1 , n ≥ 1.
Recall this means 0! = 1 and n! = n(n − 1)! for n ≥ 1. Using the recursive deﬁnition, we get:
1! = 1 · 0! = 1 · 1 = 1, 2! = 2 ...

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