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Binomial expansion I

Pascal's triangle

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Summary

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.

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

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:

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

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:

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

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 $\underline{1, \ 3, \ 3, \ 1}$ .

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:

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

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 $4^{th}$ :

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

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

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

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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Unit 1

Solving binomial problems

Unit 2

Binomial expansion

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