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Pascal's triangle (Topic)

Pascal's triangle, also called Tartaglia's triangle, is the following configuration of numbers:

\begin{displaymath} \begin{array}{cccccccccccccccccc} & & & & & & & & & 1 & & & ... ...& &\vdots & & & & \vdots & & & & \vdots& & & & \ \end{array} \end{displaymath}

In general, this triangle is constructed such that entries on the left side and right side are $1$ , and every entry inside the triangle is obtained by adding the two entries immediately above it. For instance, on the fourth row $4=1+3$ .

Historically, the application of this triangle has been to give the coefficients when expanding binomial expressions. For instance, to expand $(a+b)^4$ , one simply look up the coefficients on the fourth row, and write $$(a+b)^4 = a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4.$$

Pascal's triangle is named after the French mathematician Blaise Pascal (1623-1662) [3]. However, this triangle was known at least around 1100 AD in China; five centuries before Pascal [1]. In modern language, the expansion of the binomial is given by the binomial theorem discovered by Isaac Newton in 1665 [2]: For any $n=1,2,\ldots$ and real numbers $a,b$ , we have \begin{eqnarray*} (a+b)^n &=& \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k \\ &=& a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2} a^{n-2}b^2 + \cdots + b^n. \end{eqnarray*}Thus, in Pascal's triangle, the entries on the $n$ th row are given by the binomial coefficients $$\binom{n}{k}=\frac{n!}{(n-k)!k!}.$$ for $k=1,\ldots, n$ .

Pascal's triangle has many interesting numerical properties. For example, it is easy to see that the sum of the entries in the $n^{\mathrm{th}}$ row is $2^n$ . This can be easily proved by induction, but a more elegant proof goes as follows: $$ 2^n=(1+1)^n=\sum_{k=0}^n \binom{n}{k} 1^{n-k}1^k=\sum_{k=0}^n \binom{n}{k} $$

If you look at the long diagonals parallel to the diagonal sides of the triangle, you see in the second diagonal the integers $1,2,3,4,\ldots$ . The next diagonal down contains the triangular numbers $1,3,6,10,15,\ldots$ , and the row below that the tetrahedral number $1,4,10,20,35,\ldots$ . It is easy to see why this is: for example, each triangular number is the sum of the previous triangular number and the next integer, which precisely reflects the arrangement of the triangle. Each tetrahedral number is the sum of the previous tetrahedral number and the size of the next ``layer'' of the tetrahedron, which is just the next triangular number. Similarly, succeeding diagonals give ``triangular'' number in higher dimensions.

Bibliography

1
Wikipedia's entry on the binomial coefficients
2
Wikipedia's entry on Isaac Newton
3
Wikipedia's entry on Blaise Pascal




"Pascal's triangle" is owned by Koro. [ full author list (3) | owner history (1) ]
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See Also: binomial coefficient, Pascal's rule

Other names:  Tartaglia's triangle

Attachments:
Singmaster's conjecture (Conjecture) by PrimeFan
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Cross-references: dimensions, tetrahedron, size, reflects, tetrahedral number, triangular numbers, contains, integers, parallel, diagonals, proof, induction, sum, easy to see, properties, binomial coefficients, real numbers, binomial theorem, language, AD, Blaise Pascal, expand, expressions, binomial, coefficients, application, row, right, side, triangle, numbers, configuration
There are 19 references to this entry.

This is version 6 of Pascal's triangle, born on 2003-05-07, modified 2008-01-20.
Object id is 4248, canonical name is PascalsTriangle.
Accessed 19880 times total.

Classification:
AMS MSC05A10 (Combinatorics :: Enumerative combinatorics :: Factorials, binomial coefficients, combinatorial functions)

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proof by inspection by rspuzio on 2008-01-21 01:06:17
Looked at the right way, this fact that the sums of entries in
rows are powers of two becomes obvious. By definition, each
number in a row except the two 1's at the ends is gotten by adding
the the two elements immediately above it. So write the sum of
elements in a row, then replace the elements in the middle by
their expression as sums. For instance, rewrite

1 + 5 + 10 + 10 + 5 + 1

as

1 + 1 + 4 + 4 + 6 + 6 + 4 + 4 + 1 + 1 .

Note that each number appears twice in the latter sum. Hence,
the sum of numbers in a row equals twice the sum of the numbers in
the preceding row. Since the sum of the numbers in the top row
is 1 = 2^0, this means that the sums of the rows are successive
powers of two.
 
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