1. The Binomial
Theorem,
Fibonacci, and
Vitruvian Dan
Vitruvian Dan by flickr
user nogoodreason

2. Multiply each of the following
(x + y)0 =
(x + y)1 =
(x + y)2 =
(x + y)3 =
(x + y)4 =
(x + y)5 =

3. What's the pattern?
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

4. 1. The sum of the exponents is always equal to
the power of the binomial.
2. The exponent of the first term begins with the
same value as the power of the binomial and
decreases by one in each succesive term.
3. The exponent of the second term appears in the
second term of the expansion and increases by
one until it matches the power of the binomial.

5. 4. The number of terms is one more than the
power of the binomial.
5. The coefficients are the combinations of the
power number beginning with C(n, 0) and
ending at C(n, n) and are symmetrical.
n 0 n­2 2
(x + y)n = nC0 x y + nC1 xn­1y1 + nC2 x y + . . . . . + nCn x0yn

6. Evaluate each term ...

7. (x + y)7 =

8. (2x ­ y)4 =

9. Any individual term, let's say the ith term, in a
binomial expansion can be represented like this:
n ­ (i ­ 1) (i ­ 1)
ti = nC(i ­ 1)a b
Find the 4th term in the expansion of
(2 + x)7

10. Find the 5th term 8
( x2 ­ 2 )
x

11. Determine the indicated term in each expansion.
the 8th term in the expansion of
10
(x ­ 2 )

12. Find the term that contains x7 in the expansion of
9
(x3
­ 1
x )

13. Find the term that contains x18 in the expansion of
10
( x3 ­ 1 )
x

14. Try some more
Exercise 34, questions 1 ­ 8

15. Attachments
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Vitruvian Dan by flickr user nogoodreason