Question

Using binomial theorem, write down the expansion of the following: ${{\left( 2x+3y \right)}^{5}}$

Answer 1

Hint: We are asked to find the binomial expansion of an expression. For that we should be aware about the general binomial expansion of any expression involving two variables $x$ and $y$ raised to the power of $n$. Moreover, we should be able to find the value of the combination term involved in the binomial expansion.

Complete step by step answer:
For any two real numbers $a$ and $b$, and for any natural number $n$, the following binomial expansion holds true:
$\left(a+b\right)^n={}^nC_0a^nb^0+{}^nC_1a^{n-1}b^1+\ldots+{}^nC_na^0b^n$
So, we put $a=2x$, $b=3y$ and $n=5$, we get:
${{\left( 2x+3y \right)}^{5}}{{=}^{5}}{{C}_{0}}{{\left( 2x \right)}^{5}}{{+}^{5}}{{C}_{1}}{{\left( 2x \right)}^{4}}\left( 3y \right){{+}^{5}}{{C}_{2}}{{\left( 2x \right)}^{3}}{{\left( 3y \right)}^{2}}{{+}^{5}}{{C}_{3}}{{\left( 2x \right)}^{2}}{{\left( 3y \right)}^{3}}{{+}^{5}}{{C}_{4}}\left( 2x \right){{\left( 3y \right)}^{4}}{{+}^{5}}{{C}_{5}}{{\left( 3y \right)}^{5}}$
Now, we know that:
${}^nC_r=\dfrac{n!}{r!\left(n!-r!\right)}$
So, after dissolving the coefficients, we have:
${{\left( 2x+3y \right)}^{5}}={{\left( 2x \right)}^{5}}+5{{\left( 2x \right)}^{4}}\left( 3y \right)+10{{\left( 2x \right)}^{3}}{{\left( 3y \right)}^{2}}+10{{\left( 2x \right)}^{2}}{{\left( 3y \right)}^{3}}+5\left( 2x \right){{\left( 3y \right)}^{4}}+{{\left( 3y \right)}^{5}}$
$=32{{\left( x \right)}^{5}}+5\times 16{{\left( x \right)}^{4}}\left( 3y \right)+10{{\left( 2x \right)}^{3}}\times 9{{\left( y \right)}^{2}}+10{{\left( 2x \right)}^{2}}\times 27{{\left( y \right)}^{3}}+5\left( 2x \right)81{{\left( y \right)}^{4}}+243{{\left( y \right)}^{5}}$
$=32{{\left( x \right)}^{5}}+5\times 16{{\left( x \right)}^{4}}\left( 3y \right)+10\times 8{{\left( x \right)}^{3}}\times 9{{\left( y \right)}^{2}}+10\times 4{{\left( x \right)}^{2}}\times 27{{\left( y \right)}^{3}}+5\left( 2x \right)81{{\left( y \right)}^{4}}+243{{\left( y \right)}^{5}}$
$=32x^5+240x^4y+720x^3y^2+1080x^2y^3+810xy^4+243y^5$
Hence, the resultant expression has been obtained.

Note: Since the number to which this expression is raised is 5, which is very small; we can simply use the Pascal’s triangle to determine the coefficients of the variables. Using that we would reduce the chances of calculation mistakes that might occur while solving the combination term. The entry in the $n^{th}$ row and $k^{th}$ column of Pascal's triangle is denoted by:
$^{n}{{C}_{k}}$
The first few rows of a Pascal’s triangle are:

Exponent	Coefficients
N=1	1 1
N=2	1 2 1
N=3	1 3 3 1
N=4	1 4 6 4 1
N=5	1 5 10 10 5 1

Using this, we can write the coefficients along with the variables and then do the multiplication to obtain the resultant expression. As we can see this has been obtained only after dissolving the coefficients as we had done in the question.