Pascal’s Triangle and the Binomial Theorem, then Exam! 20.0 Students know the binomial theorem and...
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Pascal’s Triangle and the Binomial Theorem, then Exam! 20.0 Students know the binomial theorem and use it to expand binomial expressions that are raised to positive integer powers.
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Transcript of Pascal’s Triangle and the Binomial Theorem, then Exam! 20.0 Students know the binomial theorem and...
Pascal’s Triangle and the Binomial Theorem, then Exam!
20.0 Students know the binomial theorem and use it to expand binomial expressions that are raised to positive
integer powers.
Pascal’s Triangle and the Binomial Theorem
Objectives• Relate Pascal’s Triangle to
the terms of a Binomial Expansion– The Binomial Theorem
Key Words• Pascal’s triangle
– The arrangement of in a triangular pattern in which each row corresponds to a value of n. (pg 553, you have to see it to believe it!)
Pascal’s Triangle
If you arrange the values of in a triangular pattern in which each row corresponds to a value of n, you get a pattern called Pascal’s triangle.
Turn to page 553.
The Binomial Theorem
For any positive integer n, the expansion of is:
+
Note that each term has the form where r is an integer from 0 to n.
Examples:
Example 1 Expand a Power of a Simple Binomial Sum
Coefficients: 1, 4, 6, 4, 1
Powers of a: a 4, a 3, a 2, a 1, a 0
Powers of b: b0, b1, b2, b3, b4
Expand ( )4. ba +
SOLUTION
In , the power is n 4. So, the coefficients of the terms are the numbers in the 4th row of Pascal’s Triangle.
=( )4ba +
= 1a 4b 0 + + + +( )4ba + 4a 3b 1 6a 2b 2 1a 0b 44a 1b 3
= a 4 + + + +4a 3b 6a 2b 2 4ab 3 b 4
Example 2 Expand a Power of a Binomial Sum
SOLUTION
Use the binomial theorem with a x and b 5.= =
Expand ( )3. 5x +
= + + +( )x 3( )1 ( )1 ( )x 2( )3 ( )5 ( )x 1( )3 ( )25 ( )x 0( )1 ( )125
= +x 3 15x 2 75x 125+ +
5x( +
(
3 = 3C0x 350 + + +3C1x 251 3C2x 152
3C3x 053
Example 3 Expand a Power of a Binomial Difference
First rewrite the difference as a sum:
SOLUTION
= 2x[ + 4(
( – y ]y2x( – ) 4
= =Then use the binomial theorem with a 2x and b –y.= =
Expand y2x( – )4.
4C3 4C2
4C4
= + +2x[ + 4(
( – y ]4C0
(
(2x 4 (
( – y 04C1
(
(2x 3 (
( – y 1
(
(2x 2 (
( – y 2 +
(
(2x 1 (
( – y 3 +
(
(2x 0 (
( – y 4
Example 3 Expand a Power of a Binomial Difference
= + + +
+
( )16x 4( )1 ( )1 ( )8x 3( )4 ( )y– ( )4x 2( )6 ( )y 2
( )2x4 ( )– y 3 ( )1 ( )1 ( )y 4
= 16x 4 + + – +32x 3y 24x 2y 2 8xy 3 y 4
CheckpointExpand a Power of a Binomial Sum or Difference
ANSWER a 5 + + + + +5a 4b 10a 3b 2 10a 2b 3 5ab 4 b 5
ANSWER x 4 + + + +8x 3 24x 2 32x 16
Expand the power of the binomial sum or difference.
1. ba( + )5
2. 2x( + )4
3. 53x( + )3
ANSWER 27x 3 + + +135x 2 225x 125
CheckpointExpand a Power of a Binomial Sum or Difference
5. – nm( )4
6. t3s( – )3
Expand the power of the binomial sum or difference.
4. 4p( )3–
p 3ANSWER +– –12p 2 48p 64
m 4ANSWER +– – +4m 3n 6m 2n 2 4mn 3 n 4
27s 3 +– –27s 2t 9st 2 t 3ANSWER
Conclusions
SummaryExit Slip:• How can you calculate the
coefficients of the terms of ?– Each term in the expansion of
has the form , where r is an integer from 0 to n.
Assignment• Pg 555
– #(2,6-13)
• Write the assignment down, you will work on it after the you finish the exam, early.
• Get ready for the exam.
45 MINUTESNO TALKING – READ RUBRIC – READ DIRECTIONS – GOOD LUCK!
Exam on the Fundamental Counting Principle
