# How do you use the Binomial Theorem to expand (x + y)^5?

Feb 10, 2016

${\left(a + b\right)}^{5} = {a}^{5} + 5. {a}^{4.} b + 10. {a}^{3.} {b}^{2} + 10. {a}^{2.} {b}^{3} + 5. {a}^{1.} {b}^{4} + {b}^{5}$

#### Explanation:

The binomial theorem tells us that if we have a binomial (a+b) raised
to the ${n}^{t h}$ power the result will be

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} {c}_{k}^{n} \cdot {a}^{n - k} \cdot {b}^{n}$

where " "c _k^n= (n!)/(k!(n-k)!)

and is read "n CHOOSE k equals n factorial divided by k factorial (n-k) factorial".

So ${\left(a + b\right)}^{5} = {a}^{5} + 5. {a}^{4.} b + 10. {a}^{3.} {b}^{2} + 10. {a}^{2.} {b}^{3} + 5. {a}^{1.} {b}^{4} + {b}^{5}$

we notice that the powers of ' a ' keeps decreasing from 5 (which representes ' n ') until it reaches ${a}^{\text{zero}}$ at the last term.

also we notice that the power of ' b ' keeps increasing from zero untill it reaches 5 at the last term.

Now the we have to determine the coefficient of each term through the...

c_k^n= (n!)/(k!(n-k)!)

first coefficient c_0^5=(5!)/(0! .5!)=1

second c_1^5=(5!)/(1! .4!)=5

c_2^5=(5!)/(2! .3!)=10

c_3^5=(5!)/(3! .2!)=10

c_4^5=(5!)/(4! .1!)=5

c_5^5=(5!)/(5!.0!)=1

but the calculation of combinations can be tedious..so fortunately
there is an awesome way to determine the binomial coefficients which is Pascal's triangle it is easy to deduce this triangle :

hope that helps ! : )