Solve \[{{\left( 2x+3y \right)}^{5}}\] using binomial theorem.

Hint: In this type of question we have to use the concept of binomial theorem. We know that the binomial theorem is a mathematical statement that expresses for any positive integer n, the \[{{n}^{th}}\] power of the sum of two terms \[a\] and \[b\]. We will write the binomial theorem as \[{{\left( a+b \right)}^{n}}={}^{n}{{c}_{0}}{{a}^{n}}+{}^{n}{{c}_{1}}{{a}^{n-1}}b+{}^{n}{{c}_{2}}{{a}^{n-2}}{{b}^{2}}+{}^{n}{{c}_{3}}{{a}^{n-3}}{{b}^{3}}+\cdots \cdots \cdots +{}^{n}{{c}_{n-1}}a{{b}^{n-1}}+{}^{n}{{c}_{n}}{{b}^{n}}\]. The constants \[{}^{n}{{c}_{0}},{}^{n}{{c}_{1}},{}^{n}{{c}_{2}},{}^{n}{{c}_{3}},\cdots \cdots \cdots \] are called as binomial coefficients which can be obtained by using the formula \[{}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].


Complete step-by-step solution:
Now we have to expand \[{{\left( 2x+3y \right)}^{5}}\] by using the binomial theorem.
Let us consider the statement of binomial theorem which states that for any positive integer n,
\[{{\left( a+b \right)}^{n}}={}^{n}{{c}_{0}}{{a}^{n}}+{}^{n}{{c}_{1}}{{a}^{n-1}}b+{}^{n}{{c}_{2}}{{a}^{n-2}}{{b}^{2}}+{}^{n}{{c}_{3}}{{a}^{n-3}}{{b}^{3}}+\cdots \cdots \cdots +{}^{n}{{c}_{n-1}}a{{b}^{n-1}}+{}^{n}{{c}_{n}}{{b}^{n}}\] where \[{}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
If we compare the given expression i.e. \[{{\left( 2x+3y \right)}^{5}}\] with \[{{\left( a+b \right)}^{n}}\] we get,
\[\Rightarrow a=2x,b=3y\And n=5\]
Hence, by using binomial expansion stated above we can write
\[\Rightarrow {{\left( 2x+3y \right)}^{5}}={}^{5}{{c}_{0}}{{\left( 2x \right)}^{5}}+{}^{5}{{c}_{1}}{{\left( 2x \right)}^{4}}\left( 3y \right)+{}^{5}{{c}_{2}}{{\left( 2x \right)}^{3}}{{\left( 3y \right)}^{2}}+{}^{5}{{c}_{3}}{{\left( 2x \right)}^{2}}{{\left( 3y \right)}^{3}}+{}^{5}{{c}_{4}}{{\left( 2x \right)}^{1}}{{\left( 3y \right)}^{4}}+{}^{5}{{c}_{5}}{{\left( 3y \right)}^{5}}\]We know that, the binomial coefficients which are equidistant from the beginning and from the ending are of equal value i.e.\[{}^{n}{{c}_{0}}={}^{n}{{c}_{n}},{}^{n}{{c}_{1}}={}^{n}{{c}_{n-1}},{}^{n}{{c}_{2}}={}^{n}{{c}_{n-2}},\cdots \cdots \cdots \]. Hence, in this case we can write \[{}^{5}{{c}_{0}}={}^{5}{{c}_{5}},{}^{5}{{c}_{1}}={}^{5}{{c}_{4}},{}^{5}{{c}_{2}}={}^{5}{{c}_{3}}\].
Also by substituting the values \[{}^{5}{{c}_{0}}=\dfrac{5!}{0!5!}=1,{}^{5}{{c}_{1}}=\dfrac{5!}{1!4!}=\dfrac{5\times 4!}{1!4!}=5,{}^{5}{{c}_{2}}=\dfrac{5!}{2!3!}=\dfrac{5\times 4\times 3!}{\left( 2\times 1 \right)\times 3!}=10\] which we have calculated by using the formula \[{}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] and \[n!=n\times \left( n-1 \right)!=n\times \left( n-1 \right)\times \left( n-2 \right)!=n\times \left( n-1 \right)\times \left( n-2 \right)\times \cdots \cdots \times 2\times 1\] we can write the above expansion
\[\Rightarrow {{\left( 2x+3y \right)}^{5}}=1{{\left( 2x \right)}^{5}}+5{{\left( 2x \right)}^{4}}\left( 3y \right)+10{{\left( 2x \right)}^{3}}{{\left( 3y \right)}^{2}}+10{{\left( 2x \right)}^{2}}{{\left( 3y \right)}^{3}}+5{{\left( 2x \right)}^{1}}{{\left( 3y \right)}^{4}}+1{{\left( 3y \right)}^{5}}\]
By simplifying it further we get
\[\Rightarrow {{\left( 2x+3y \right)}^{5}}={{2}^{5}}{{x}^{5}}+\left( 5\centerdot {{2}^{4}}\centerdot 3\centerdot {{x}^{4}}\centerdot y \right)+\left( 10\centerdot {{2}^{3}}\centerdot {{3}^{2}}\centerdot {{x}^{3}}\centerdot {{y}^{2}} \right)+\left( 10\centerdot {{2}^{2}}\centerdot {{3}^{3}}\centerdot {{x}^{2}}\centerdot {{y}^{3}} \right)+\left( 5\centerdot 2\centerdot {{3}^{4}}\centerdot x\centerdot {{y}^{4}} \right)+{{3}^{5}}{{y}^{5}}\]\[\begin{align}
  & \Rightarrow {{\left( 2x+3y \right)}^{5}}=32{{x}^{5}}+\left( 5\centerdot 16\centerdot 3\centerdot {{x}^{4}}\centerdot y \right)+\left( 10\centerdot 8\centerdot 9\centerdot {{x}^{3}}\centerdot {{y}^{2}} \right)+\left( 10\centerdot 4\centerdot 27\centerdot {{x}^{2}}\centerdot {{y}^{3}} \right)+\left( 5\centerdot 2\centerdot 81\centerdot x\centerdot {{y}^{4}} \right)+243{{y}^{5}} \\
 & \Rightarrow {{\left( 2x+3y \right)}^{5}}=32{{x}^{5}}+\left( 240{{x}^{4}}y \right)+\left( 720{{x}^{3}}{{y}^{2}} \right)+\left( 1080{{x}^{2}}{{y}^{3}} \right)+\left( 810x{{y}^{4}} \right)+243{{y}^{5}} \\
 & \Rightarrow {{\left( 2x+3y \right)}^{5}}=32{{x}^{5}}+240{{x}^{4}}y+720{{x}^{3}}{{y}^{2}}+1080{{x}^{2}}{{y}^{3}}+810x{{y}^{4}}+243{{y}^{5}} \\
\end{align}\]Hence, by using binomial theorem the expansion of \[{{\left( 2x+3y \right)}^{5}}\] is \[32{{x}^{5}}+240{{x}^{4}}y+720{{x}^{3}}{{y}^{2}}+1080{{x}^{2}}{{y}^{3}}+810x{{y}^{4}}+243{{y}^{5}}\].

Note: In this type of question students have to remember the binomial theorem. Students have to note that to find \[{}^{n}{{c}_{0}},{}^{n}{{c}_{1}},{}^{n}{{c}_{2}},{}^{n}{{c}_{3}},\cdots \cdots \cdots \] they have to use the formula \[{}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]. Also in calculation of \[{}^{n}{{c}_{0}},{}^{n}{{c}_{1}},{}^{n}{{c}_{2}},{}^{n}{{c}_{3}},\cdots \cdots \cdots \] students have to remember to use the formula \[n!=n\times \left( n-1 \right)!=n\times \left( n-1 \right)\times \left( n-2 \right)!=n\times \left( n-1 \right)\times \left( n-2 \right)\times \cdots \cdots \times 2\times 1\]. Students have to take care in calculating the value of \[{}^{5}{{c}_{0}}\] as it includes \[0!\] and they have to remember that \[0!=1\].