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# Pascal’s triangle

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

$\begin{array}[]{cccccccccccccccccc}&&&&&&&&&1&&&&&&&&\\ &&&&&&&&1&&1&&&&&&&\\ &&&&&&&1&&2&&1&&&&&&\\ &&&&&&1&&3&&3&&1&&&&&\\ &&&&&1&&4&&6&&4&&1&&&&\\ &&&&1&&5&&10&&10&&5&&1&&&\\ &&&1&&6&&15&&20&&15&&6&&1&&\\ &&1&&7&&21&&35&&35&&21&&7&&1&\\ &&&&&\vdots&&&&\vdots&&&&\vdots&&&&\\ \end{array}$ |

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

$(a+b)^{4}=a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4}.$ |

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

$\displaystyle(a+b)^{n}$ | $\displaystyle=$ | $\displaystyle\sum_{{k=0}}^{n}\binom{n}{k}a^{{n-k}}b^{k}$ | ||

$\displaystyle=$ | $\displaystyle a^{n}+\binom{n}{1}a^{{n-1}}b+\binom{n}{2}a^{{n-2}}b^{2}+\cdots+b% ^{n}.$ |

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.

# References

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

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## Corrections

A plus sign by pahio ✓

some minor stuff by GrafZahl ✓

Add synonym Tartaglia's Triangle by kfgauss70 ✓

## Comments

## proof by inspection

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.

## Re: proof by inspection

Isn't that just the induction argument alluded to in the entry?

Another slick proof is to do it combinatorially: The k-th entry in the n-th row is just the number of ways of choosing k out of n objects. The sum of all the entries in the row is the number of ways of choosing any subset of the n objects, of which there are 2^n.

Cam

## Re: proof by inspection

> Isn't that just the induction argument alluded to in the entry?

Pretty much, just presented in a way which can be easily grasped

by a fifth-grader. Personally, I don't see what is so inelegant

about the induction proof --- especially as presented here, it

follows rather immediately from the construction of the triangle.

For me, it is less elegant to first prove the binomial theorem,

but this is a matter of taste so it is not like either me or Koro

are right or wrong. At any rate, nice to have three proofs ---

now, I have no doubts about this result ;)

## Re: proof by inspection

Sure -- I certainly wasn't arguing that your proof (or the more formal induction method) was any less elegant than any other method.

Cam

## Re: proof by inspection

> Sure -- I certainly wasn't arguing that your proof (or the

> more formal induction method) was any less elegant than any

> other method.

The comment wasn't directed at you --- rather it is a response

to what the entry says --- "This can be easily proved by

induction, but a more elegant proof goes as follows:". However,

as I said earlier, elegance is a matter of personal taste, so

you, Koro, and I are equally entitled to our opinions

as to elegance.

## Re: proof by inspection

correction: it should have said

"you Koro, Matte, and I" since Matte is a coauthor.

In addition to being able to search fora, it would be

nice to be able to edit posts.