PASCAL'S TRIANGLE & the BINOMIAL THEOREM

Introduction

Blaise Pascal, (1623 - 1662) like Descartes and other French thinkers through the ages, was not only a mathematician, but a philosopher, physicist, author and social critic as well. His work in math includes a time-saving tool known as Pascal's triangle which we still use today to determine the coefficients in the expansion of ( a + b ) n where n is a positive integer.

The values in the triangular table are n C r or , the number of combinations we can make from n objects taken r at a time for r between 0 and n.

Pascal's Triangle

We'll use the expansion of ( a + b ) ³ to analyze the setup.

( a + b ) ³ = a³ + 3a²b + 3ab² + b³

We see that the powers of a descend from 3 to 0 while the
powers of b ascend from 0 to 3. (Recall that a 0 = 1).

So the expansion of ( a + b ) n = a n + __a n - 1 b + __a n - 2 b² + __a n - 3 b³ + ...... + b n .

Pascal's Triangle fills the blanks -- it lists the coefficients of the terms in the polynomial. Row 1 says ( a + b ) 1 = 1a + 1b
Row 2 says ( a + b ) ² = 1a² + 2ab + 1b²
Row 3 says ( a + b ) ³ = 1a³ + 3a²b + 3ab² + 1b³

Therefore the nth row would say
( a + b ) n = n C 0 a n b 0 + n C 1 a n - 1 b 1 + n C 2 a n - 2 b² + ...... + n C n a 0 b n.

This is known as the BINOMIAL THEOREM

So from the table, ( a + b ) 5 = 1a 5 + 5a4b + 10a³b² + 10a²b³ + 5ab 4 + 1b 5.

If we set b = 4, then ( a + 4 ) ³ = 1a³ + 3a²(4) + 3a(4)² + 1(4)³.

Since ( a - 4 ) = [ a + ( - 4 )] we expand setting b = - 4.

This theorem can be used to expand any binomial such as .

Sigma Notation for a Binomial Series

The expansion of ( a + b ) n is a sum of ( n + 1 ) terms, so we can use summation notation.  When we have , we notice that the a n - r term is missing. We have only b r.

So we rewrite it , setting a = 1, since 1 to any power equals 1.

For example, if we see we rewrite it as ( 1 + cos x )³.

Solving Equations

Examples:

1) Solve for x if Since , we have ( x + 3 ) 8 = 0 so x = -3.

2) Solve for x if Since , we have ( x - 4 ) 5 = 2 5 so x = 6.

Note: ( -1 ) r with r from 0 to n is positive 1 when r is even and negative 1 when r is odd.
So all it does is change the sign of the terms.

This is known as an alternating series.

With the binomial thorem, we combine this term with the second variable
to get ( a - b )n instead of ( a + b ) n .

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Constructing the Table Notice that each row begins and ends with 1 -- since n C 0 and n C n both = 1

Each row has n + 1 entries.

From Row 2 on down, each central entry is the sum of the 2 entries above it.

Row 6 would read:

 1 = 6 C 0 6 = 6 C 1 15 = 6 C 2 20 = 6 C 3 15 = 6 C 4 6 = 6 C 5 1 = 6 C 6

The values are symmetric because n C r = n C n - r .
In words: choosing 1 of 6 is the same as leaving 5 of 6.

calculator hint: on the TI-83 series we find the combinations functions in the MATH menu. We enter the value for n, then choose MATH -- PRB -- #3 for combinations -- then enter the value for r. We can do it with the CATALOG menu, but it's slower.

efficiency hint: when finding a lot of them, find the first one as above, then,
hit "entry" (2nd-ENTER) to display the last entry. Scroll in and change the values of n and r.
By the way, the 10 last entries can be retrieved this way.

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Examples:

1) Expand (2t - 5)³.

We set a = 2t, and b = -5. We use the entries in row 3 since n = 3.

(2t - 5) ³ = 1(2t) ³ + 3(2t) ²(-5) + 3(2t)(-5) ² + 1(-5) ³

= 8t ³ - 60t ² + 150t - 125

2) Find the 5th term of (2x - 5)6 .

We set a = 2x, and b = -5. We use the 5th entry in row 6 since n = 6.

Because we start at 0, the 5th entry is 6 C 4 . (Use the table above for the 6th row.)

The 5th term is 6 C 4 (2x)²(-5)4 = 15(4x²)(625) = 37, 400 x².

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Practice

1) Expand the binomial series:

 a) ( a - 2b) 4 . b) ( a² - 3b) 5 . c) ( 2a - 3b² )³

2) Find the indicated term of the binomial expression:

 a) 3rd term of ( a + b) 6 . b) 6th term of ( x + y) 7 . c) 12th term of ( a - 2 ) 14 .

3) Find the middle term of the expansion of ( 2a - b ) 12.

4) Solve for x:

 a) if b) if ; x ` · (radians)

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Solutions

1) Expand the binomial series:

 a) ( a - 2b) 4 = a 4 - 8a³b + 24a² b² - 32ab³ + 16b 4 . b) ( a² - 3b) 5 = a 10 - 15a 8 b + 90a 6 b² - 270a 4 b³ + 405a² b 4 - 243b 5 . c) ( 2a - 3b² )³ = 8a³ - 36a² b² + 54ab 4 - 27b 6 .

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2) Find the indicated term of the binomial expression:

 a) 3rd term of ( a + b) 6 is 15a 4 b² b) 6th term of ( x + y) 7 is 21x² y 5 . c) 12th term of ( a - 2 ) 14 is 745, 472 a³

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3) Since n = 12, there are 13 terms so the 7th term is the middle term.

The 7th term = 59, 136 b 6 a 6 .

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4) a) we have ( 5 + x ) 4 = 64, so we take the 4th root of both sides to get: b) we have ( 1 + sin x ) 9 = 0, so we take the 9th root of both sides to get:

1 + sin x = 0, which makes sin x = - 1

Since sin x = - 1, x must be all Real multiples of The solution is .

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