Pascal's triangle

In a nutshell

Pascal's Triangle provides you with a quick and easy method to read off the coefficients of the terms in the expansion of $$(x + y)^n$$. In particular, the $$m^{th}$$ entry of the $$n^{th}$$ row is the coefficient of the $$x^{n-m}y^{m-1}$$ term in the expansion of $$(x+y)^{n-1}$$. The entries of the $$n^{th}$$ row of Pascal's Triangle are easily obtained from those of the $$(n-1)^{th}$$ row, making it easy to reproduce the whole triangle at a moment's notice.

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​$$\begin{array}{ccccccc}&&&&1&&&\\&&&1&&1&&\\&&1&&2 &&1&\\&1&&3&&3&&1\end{array}$$​​

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Computing entries

Row $$n$$ has $$n$$ entries, and the first and last entry of every row is given by a $$1$$​. To compute the entry directly below two others, simply add the two together.

Example 1

Compute the $$4^{th}$$ row of Pascal's Triangle.

The rules above tell you that the first two rows must be given by:

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$$\begin{array}{ccc}&&1&\\&1&&1\\\end{array}$$

The first and last entry of row $$3$$​ is $$1$$. To obtain the second entry of row $$3$$, compute the sum of the first and second entries in the second row as $$1+1=2$$, and hence find that the triangle now looks like this:

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$$\begin{array}{ccccc}&&&1&&\\&&1&&1&\\&1&&2&&1\end{array}$$​​

The first and last entry of row $$4$$ is $$1$$. To obtain the second entry of the fourth row, add together the first two entries of row $$3$$ to obtain $$1 + 2 = 3$$. Similarly, to obtain the third entry in row $$4$$, add together the second and third entries of row $$3$$ to obtain $$2+1=3$$. The triangle now looks like this:

​$$\begin{array}{ccccccc}&&&&1&&&\\&&&1&&1&&\\&&1&&2 &&1&\\&1&&3&&3&&1\end{array}$$​​

Therefore, the fourth row of Pascal's triangle is $$\underline{1, \ 3, \ 3, \ 1}$$.

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Relation to binomial expansions

It can be tedious to carry out the full expansion of $$(x+y)^n$$ for large values of $$n$$ - however, Pascal's triangle provides you with a convenient shortcut! Observe:

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​$$\begin{array}{cccccccccccc}&(x+y)^{0}&&=&&&&&&1&&&\\&(x+y)^{1}&&=&&&&&1x&+&1y&&\\&(x+y)^{2}&&=&&&&1x^2&+&2xy&+&1y^2&\\&(x+y)^{3}&&=&&&1x^3&+&3x^2y&+&3xy^2&+&1y^3\end{array}$$​​

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This pattern continues, hence reducing the problem of computing the expansion of $$(x+y)^n$$​ to that of producing row $$n$$ of Pascal's triangle. It follows that the $$m^{th}$$​ entry of the $$n^{th}$$​ row is the coefficient of the $$x^{n-m}y^{m-1}$$​ term in the expansion of $$(x+y)^{n-1}$$​.​

Note: The exponents of $$x$$​ and $$y$$​ in the terms in the $$n^{th}$$ row of the triangle should sum to $$n-1$$, and we enter them so that the powers of $$x$$ are descending and the powers of $$y$$ are ascending.

Example 2

Expand $$(x+y)^4$$.

Following the working from Example 1, we compute the $$5^{th}$$ row of Pascal's triangle via the entries from the $$4^{th}$$:

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​$$1 + 3 = 4 \\3 + 3 = 6 \\3 + 1 = 4$$

Thus the triangle up to row $$4$$ is given by:

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$$\begin{array}{ccccccccc}&&&&&1&&&&\\&&&&1&&1&&&\\&&&1&&2 &&1&&\\&&1&&3&&3&&1&\\&1&&4&&6&&4&&1\end{array}$$​​

Using the guide for assigning entries as coefficients from above, observe that powers of $$x$$ should be descending from left to right, the powers of $$y$$ should be ascending, and the sums of the powers of $$x$$ and $$y$$ in each term should sum to $$5-1 = 4$$.

Therefore, $$\underline{( x+y)^4 = x^4 + 4x^3y +6x^2y^2 + 4xy^3 + y^4}$$.

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