Question

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

Answer 1

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{aligned} & \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{aligned}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 $0!=1$.