Question

How do you use the binomial theorem to expand ${\left( {x + y} \right)^5}$?

Answer 1

Hint: The above question is based on the concept of binomial theorem. The main approach towards solving this question is by expanding the expression using the binomial theorem where there is expansion of powers of the binomial which results in the sum of all the terms.

Complete step by step solution:
In mathematics the binomial expansion describes the algebraic expansion of the powers of a binomial.
Binomial here is a polynomial with two terms. Here the two terms are the variable x and the variable y.
In the above expression we can see that the polynomial is multiplied by itself multiple times. That means the polynomial is multiplied by itself five times since the power of the expression is 5.
Now the calculations get longer as we go on multiplying similar terms and therefore some kind of pattern is developing and this is called as binomial theorem.
Now writing in general terms we get,
\[{\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {c_k^n \times {a^{n - k}} \times {b^n}} \]
where \[c_k^n = \dfrac{{n!}}{{k!\left( {n - k} \right)!}}\]
Now here the value of n is 5 and the value of a is x and b is y. So, substituting and expanding it we get,
\[
c_0^5 = \dfrac{{5!}}{{0!5!}} = 1 \\
\Rightarrow c_1^5 = \dfrac{{5!}}{{1!4!}} = 5 \\
\Rightarrow c_2^5 = \dfrac{{5!}}{{2!3!}} = 10 \\
\Rightarrow c_3^5 = \dfrac{{5!}}{{3!2!}} = 10 \\
\Rightarrow c_3^5 = \dfrac{{5!}}{{3!2!}} = 10 \\
\Rightarrow c_4^5 = \dfrac{{5!}}{{4!1!}} = 5 \\
\Rightarrow c_5^5 = \dfrac{{5!}}{{5!0!}} = 1 \\ \]
So, in the above we can find the coefficient of each term. We get the following terms,
\[{\left( {x + y} \right)^5} = {x^5} + 5{x^4}y + 10{x^3}{y^2} + 10{x^2}{y^3} + 5x{y^4} + {y^5}\]

Note: An important thing to note is that the powers of each term keep decreasing until it reaches the last term. The pattern of the powers is written in such a way the two variables powers addition should be a sum of 5. For example 3+2=5,5+0=5,1+4=5,etc.