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

**Pascal's Triangle**

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

( *a + b* ) ³ = *a*³ + 3*a*²*b* + 3*ab*² + *b*³

We see that the **powers of a descend** from 3 to 0 while the

So the expansion of ( *a + b* )^{ n} = *a*^{ n} + __*a*^{ n - 1} *b* + __*a*^{ n - 2} *b*² + __*a*^{ n - 3} *b*³ + ...... + *b*^{ n} .

Pascal's Triangle fills the blanks -- it lists the **coefficients** of the terms in the polynomial.

Row 1 says ( *a + b* )^{ 1} = 1*a* + 1*b*

Row 2 says ( *a + b* ) ² = 1*a*² + 2*ab* + 1*b*²

Row 3 says ( *a + b* ) ³ = 1*a*³ + 3*a*²*b* + 3*ab*² + 1*b*³

Therefore the *n*th 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} *b*² + ...... + _{n }C_{ n}*a*^{ 0} *b *^{n}.

This is known as the **BINOMIAL THEOREM**

So from the table, ( *a + b* )^{ 5} = 1*a*^{ 5} + 5*a*^{4}*b* + 10*a*³*b*² + 10*a*²*b*³* + *5*ab*^{ 4} + 1*b*^{ 5}.

If we set *b* = 4, then ( *a + *4 ) ³ = 1*a*³ + 3*a*²(4) + 3*a*(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

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**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

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}** =

In words:

**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 +

= 8t ³ - 60t ² + 150t - 125

2) Find the 5th term of (2*x* - 5)^{6} .

We set *a* = 2*x*, 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} (**2

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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 ( 2*a* - *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} = a4^{ } - 8a³b + 24a² b² - 32ab³ + 16b4^{ } . |

b) ( a² - 3b)^{ 5} = a10^{ } - 15a8^{ } b + 90a6^{ } b² - 270a4^{ } b³ + 405a² b4^{ } - 243b5^{ } . |

c) ( 2a - 3b² )³ = 8a³ - 36a² b² + 54ab4^{ } - 27b6^{ } . |

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2) Find the indicated term of the binomial expression:

a) 3rd term of ( a + b)^{ 6} is 15 |
b) 6th term of ( x + y)^{ 7} is 21 |
c) 12th term of ( a - 2 )^{ 14} is 745, 472 |

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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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