**Unformatted text preview: **and we note that P (1) is true since
1
? (a + b)1 =
j =0
? a+b = 1 1−j j
ab
j 1 1−0 0
1 1−1 1
a b+
ab
0
1 a+b = a+b
Now we assume that P (k ) is true. That is, we assume that we can expand (a + b)k using the
formula given in Theorem 9.4 and attempt to show that P (k + 1) is true.
(a + b)k+1 = (a + b)(a + b)k
k = (a + b)
j =0
k =a
j =0
k =
j =0 k k−j j
ab
j k k−j j
a b +b
j
k k+1−j j
a
b+
j k
j =0
k
j =0 k k−j j
ab
j
k k−j j +1
ab
j Our goal is to combine as many of the terms as possible within the two summations. As the counter
j in the ﬁrst summation runs from 0 through k , we get terms involving ak+1 , ak b, ak−1 b2 , . . . , abk .
In the second summation, we get terms involving ak b, ak−1 b2 , . . . , abk , bk+1 . In other words, apart
from the ﬁrst term in the ﬁrst summation and the last term in the second summation, we have
4 and a fair amount of tenacity and attention to detail. 586 Sequences and the Binomial Theorem terms common to both summations. Our next...

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