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**Unformatted text preview: **and n = 3, we get
1
2+
10 3 3 =
j =0 = 3 3−j
2
j 3 3−0
2
0 = 8+ 1
10 1
10
0 + j 3 3−1
2
1 12
6
1
+
+
10 100 1000 = 8 + 1.2 + 0.06 + 0.001
= 9.261 1
10 1 + 3 3−2
2
2 1
10 2 + 3 3−3
2
3 1
10 3 588 Sequences and the Binomial Theorem 3. Identifying a = 2x, b = y and n = 5, the Binomial Theorem gives
5
5 (2x + y ) =
j =0 5
(2x)5−j y j
j Since we are concerned with only the term containing x3 , there is no need to expand the
entire sum. The exponents on each term must add to 5 and if the exponent on x is 3, the
exponent on y must be 2. Plucking out the term j = 2, we get
5
(2x)5−2 y 2 = 10(2x)3 y 2 = 80x3 y 2
2 We close this section with Pascal’s Triangle, named in honor of the mathematician Blaise Pascal.
Pascal’s Triangle is obtained by arranging the binomial coeﬃcients in the triangular fashion below.
0
0
1
0
2
0
3
0
4
0 1
1
2
1 3
1
4
1 2
2
3
2 4
2
.
.
. 3
3
4
3 4
4 Since n = 1 and n = 1 for all whole numbers n, we get that each row of Pascal’s Triangle
0
n
begins and ends with 1. To generate the...

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