Page 143 - 'Blast_Into_Math

Blast into Math! Mathematical perspectives: all aour mase are melong to us

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Since the first sum has powers of a from a up to a n+1 , there is only one term with a power
of a n+1 and it is the last term (the term with k = n ) in the first sum :

n!
b
a n+1 0 = a n+1 .
n!

This is the same as the term with k = n +1 in:

n+1
k n−k (n +1)!
a b .
k!(n +1 − k)!
k=0

Now we just need to figure out all the terms in the middle. The remaining terms are :

n−1 n
k+1 n−k n! j n−j+1 n!
a b + a b .
k!(n − k)! j!(n − j)!
k=0 j=1

We can match up the terms that have the same power. If the power of a and b in the first
sum are

a k+1 n−k ,
b

then this corresponds to the term with j = k +1 in the second sum because this term is

n! n!
a k+1 n−(k+1)+1 = a k+1 n−k .
b
b
(k +1)!(n − (k +1))! (k +1)!(n − k − 1)!
So we can put these two together:

n! n!
b
b
a k+1 n−k + a k+1 n−k
k!(n − k)! (k +1)!(n − k − 1)!

n! n!
b
= a k+1 n−k + .
k!(n − k)! (k +1)!(n − k − 1)!
Now, remember that factorials are like onions, and (k +1)! is just k! with one more layer,

(k +1)! =(k +1)k!.

In case you haven’t already done so, it might be a good time to review your work on # 9 from
Chapter 4.

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