Question

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

Answer 1

The final answer :
`(a+b)^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^(th)` power the result will be

`(a+b)^n=sum_(k=0)^nc_k^n *a^(n-k)*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 `(a+b)^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^("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 ! : )