On this page we are mainly concerned with Pascal's Triangle expanded into negative integers. Other expansions include fractions.

On this page, I sometimes write as B(n,k)

See also Pascal's Triangle- Binomial Expansions for Negative n
- List of Expansions of Binomials with Negative n
- Pascals Extended Triangle for Negative n
- Upper Negation or Negating the Upper Index
- Proof of Upper Negation or Negating the Upper Index

Of course, we rememeber that by binomial we refer to
(a+b)^{n}, and the like. We usually expand these using the binomial **
theorem**. If we did not know the binomial theorem, then we would
expand these binomials in some other way (such as straight division). On this page, we are interested in
binomials of the form (1+x)^{−}n
where n is a positive integer or zero and |x|<1. Unless |x|<1 the
binomial does not converge for negative n. (See
binomialProofNegativeIntegers).
We explore some of these expansions which we discover by straight division
(because we have forgotten the binomial theorem, and also because we do not
want to assume what we seek to prove).

The series below are expansions for various binomials, to eleven terms and for eleven values of n:

[1.01]

All these (except when n=0) are infinite series that converge when |x|<1.

Using the information from the expansions above [1.01], derived by straight division, we can make a table as below:

Pascal's Extended Triangle | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|

k | |||||||||||

n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

-10 | 1 | -10 | 55 | -220 | 715 | -2002 | 5005 | -11440 | 24310 | -48620 | 92378 |

-9 | 1 | -9 | 45 | -165 | 495 | -1287 | 3003 | -6435 | 12870 | -24310 | 43758 |

-8 | 1 | -8 | 36 | -120 | 330 | -792 | 1716 | -3432 | 6435 | -11440 | 19448 |

-7 | 1 | -7 | 28 | -84 | 210 | -462 | 924 | -1716 | 3003 | -5005 | 8008 |

-6 | 1 | -6 | 21 | -56 | 126 | -252 | 462 | -792 | 1287 | -2002 | 3003 |

-5 | 1 | -5 | 15 | -35 | 70 | -126 | 210 | -330 | 495 | -715 | 1001 |

-4 | 1 | -4 | 10 | -20 | 35 | -56 | 84 | -120 | 165 | -220 | 286 |

-3 | 1 | -3 | 6 | -10 | 15 | -21 | 28 | -36 | 45 | -55 | 66 |

-2 | 1 | -2 | 3 | -4 | 5 | -6 | 7 | -8 | 9 | -10 | 11 |

-1 | 1 | -1 | 1 | -1 | 1 | -1 | 1 | -1 | 1 | -1 | 1 |

0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

The table above is Pascal's Extended Triangle for negative numbers. Other extensions involve fractions, etc.

We note that whenever k<n, then B(n,k)=0, that is:

[2.01

So if we consider B(0,0), which is 1 in the table, by applying the Addition Law, we note:

[2.02]

We note that the left-hand-side is 1, and B(-1,0) is one, so B(-1,-1) must be 0, as indicated in the general definition [2.01]

We can verify that the addition law seems to apply for some negative n.

Just like Pascal's Triangle, we can discover any number
from this table from the numbers in earlier rows, using the relationship:

from pascal's
triangle

First we claim to prove this formula for n as a positive integer and for n equal to zero.

If n is a positive integer then we can interpret the first binomial, n over k, as a combination, n choose k. Suppose we have n marbles, one of them is blue, and the rest are red. We can pick k marbles (containing a blue one, or not) in ways. We can select k marbles with all of them red, in ways. That is we remove the blue marble (so it cannot be selected), and pick out k red marble at random. This is all the combinations except those containing exactly one blue marble. We can select these remaining combinations by picking out the blue marble and selecting (k-1) marbles from the (n-1) red marbles to make up the k required in ways. Clearly, the total selections are:

which is the addition law. Hence the addition law is true for all positive n. (It is also true at zero, because then all the terms are zero). We also note that both sides of the equation are polynomials of degree k. By the polynomial argument, two polynomials of degree k can agree at most at k points, otherwise they are the same polynomial. The equation agrees at an infinite number of points (when n is a positive integer) hence the polynomials are identical and therefore true for all real (and complex values).

The addition law is, therefore, true for all values of n. Hence, we ought to write n as r to indicate we are referring to all numbers:

■ [3.01]

Upper Negation is also called negating the upper index.

The diagonals in Pascal's Triangle seem to contain the same numbers as the rows in Pascal's Extended Triangle (for Negative Integers) above.

There is a rule for the relationship:

[4.01]

Where r is a real number (also complex) and k is an integer. If k<0, then both sides are zero.

This is called Upper Negation, because we compose the right-hand-side binomial (after writing (-1)k) by writing down k (in the upper and lower indexes) and take the upper index, r, of the left-hand-side, negate it, (-r) and add it to the k in the upper index and finally subtract 1.

This formula is useful in solving problems involving Binomials, with no restrictions on r and for k as an integer. For instance, starting with the right-hand-side, it enables us to remove the k from the upper index, sometimes simplifying the expression.

We wish to prove Upper Negation:

[4.01, repeated]

To begin with we take r as a positive integer (or zero) and expand as a combination.

We expand the left-hand-side:

[5.01]

Now multiply every factor by (-1), k times:

[5.02]

Reverse the order of each of the factors:

[5.03]

We notice that the factors are the binomial coefficient, , so we can write:

■ [5.04]

The two sides are the same for all r as a positive integer (or zero) and for k as an integer. If k<0, then the expressions are zero by definition. As with the addition law, we can note that both sides of the equation are polynomials of degree r. Such polynomials can agree at at most r points, but as these agree at an infinite number (the positive integers) then the two sides are identical and are true for all values of r. (k must be an integer).

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