College Algebra, 2013a

662 Sequences and the Binomial Theorem
Again, the reader is encouraged to write out the sum and compare it to Definition 8.9. Our next
example gives us practice with this new notation.
Example 9.2.1.
1. Find the following sums.
(a)
4∑
k=1
13
100 k (b)
4∑
n=0
n!
2
(c)
5∑ (−1) n+1
(x − 1) n
n
n=1
2. Write the following sums using summation notation.
Solution.
(a) 1+3+5+...+ 117
(b) 1 − 1 2 + 1 3 − 1 4 + − ...+ 1
117
(c) 0.9+0.09 + 0.009 + ...0. 0 } ···0 {{ } 9
n − 1 zeros
1. (a) We substitute k = 1 into the formula 13
100 k and add successive terms until we reach k =4.
4∑
k=1
13
100 k = 13
100 1 + 13
100 2 + 13
100 3 + 13
100 4
= 0.13 + 0.0013 + 0.000013 + 0.00000013
= 0.13131313
(b) Proceeding as in (a), we replace every occurrence of n with the values 0 through 4. We
recall the factorials, n! as defined in number Example 9.1.1, number6 and get:
4∑
n=0
n!
2
= 0!
2 + 1!
2 + 2!
2 + 3!
2 = 4!
2
= 1 2 + 1 2 + 2 · 1
2 + 3 · 2 · 1 + 4 · 3 · 2 · 1
2 2
= 1 2 + 1 2 +1+3+12
= 17
(c) We proceed as before, replacing the index n, but not the variable x, with the values 1
through 5 and adding the resulting terms.