Notes: `(10x - 3y)^12 , n = 10` Find the nth term in the expansion of the binomial.

Sunday, February 13, 2011

$\displaystyle{{\left({10}{x}-{3}{y}\right)}}^{{12}},{n}={10}$ Find the nth term in the expansion of the binomial.

In order to solve this problem we can use the following equation of the binomial theorem:

$\displaystyle{{\left({a}+{b}\right)}}^{{n}}={\sum_{{{k}={1}}}^{{n}}}{\left(\frac{{{n}!}}{{{\left({n}-{k}\right)}!\cdot{k}!}}\right)}\cdot{{a}}^{{{n}-{k}}}\cdot{{b}}^{{k}}$

where:

a= first term

b= last term

n = exponent (power in original equation)

k = term required - 1

This will be clearer by solving the above example:

Our example is as follows:

$\displaystyle{{\left({10}{x}-{3}{y}\right)}}^{{12}}$

In this example we are looking for the 10th term, therefore:

$\displaystyle{a}={10}{x}$

$\displaystyle{b}={3}{y}$

$\displaystyle{n}={12}$

$\displaystyle{k}={10}-{1}={9}$ (the 10th term we are looking for so we subtract one from it)

Because we are looking for the 10th term the following equation will be used:

$\displaystyle{t}_{{{k}+{1}}}=\frac{{{n}!}}{{{\left({n}-{k}\right)}!\cdot{k}!}}\cdot{{a}}^{{{n}-{k}}}\cdot{{b}}^{{k}}$

$\displaystyle{t}_{{{9}+{1}}}=\frac{{{12}!}}{{{\left({12}-{9}\right)}!{\left({9}\right)}!}}{{\left({10}{x}\right)}}^{{{12}-{9}}}\cdot{{\left({3}{y}\right)}}^{{9}}$