Question

How do you use the Binomial Theorem to expand ${{\left( 5x+2y \right)}^{6}}$ ?

Answer 1

To expand ${{\left( 5x+2y \right)}^{6}}$ , we will use Binomial Theorem which states that ${{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}}$ , where $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ . By using this theorem, we can write ${{\left( 5x+2y \right)}^{6}}=\sum\limits_{r=0}^{6}{^{6}{{C}_{r}}{{\left( 5x \right)}^{6-r}}{{\left( 2y \right)}^{r}}}$ . First, we have to expand the summation and then apply $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ . On further simplification, we will get the required answer.

Complete step-by-step solution:
We need to expand ${{\left( 5x+2y \right)}^{6}}$ using Binomial Theorem. Let us see what Binomial Theorem is. Binomial Theorem states that
${{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}}...(i)$ , where $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
We are given that ${{\left( 5x+2y \right)}^{6}}$ . When we compare this with the Binomial Theorem, we can see that $a=5x,b=2y,n=6$ . Now, let us substitute these values in (i). We will get
${{\left( 5x+2y \right)}^{6}}=\sum\limits_{r=0}^{6}{^{6}{{C}_{r}}{{\left( 5x \right)}^{6-r}}{{\left( 2y \right)}^{r}}}$
Let us expand this.
${{\left( 5x+2y \right)}^{6}}{{=}^{6}}{{C}_{0}}{{\left( 5x \right)}^{6-0}}{{\left( 2y \right)}^{0}}{{+}^{6}}{{C}_{1}}{{\left( 5x \right)}^{6-1}}{{\left( 2y \right)}^{1}}{{+}^{6}}{{C}_{2}}{{\left( 5x \right)}^{6-2}}{{\left( 2y \right)}^{2}}{{+}^{6}}{{C}_{3}}{{\left( 5x \right)}^{6-3}}{{\left( 2y \right)}^{3}}{{+}^{6}}{{C}_{4}}{{\left( 5x \right)}^{6-4}}{{\left( 2y \right)}^{4}}{{+}^{6}}{{C}_{5}}{{\left( 5x \right)}^{6-5}}{{\left( 2y \right)}^{5}}{{+}^{6}}{{C}_{6}}{{\left( 5x \right)}^{6-6}}{{\left( 2y \right)}^{6}}$
Let us simplify this further.
${{\left( 5x+2y \right)}^{6}}{{=}^{6}}{{C}_{0}}{{\left( 5x \right)}^{6}}{{+}^{6}}{{C}_{1}}{{\left( 5x \right)}^{5}}2y{{+}^{6}}{{C}_{2}}{{\left( 5x \right)}^{4}}{{\left( 2y \right)}^{2}}{{+}^{6}}{{C}_{3}}{{\left( 5x \right)}^{3}}{{\left( 2y \right)}^{3}}{{+}^{6}}{{C}_{4}}{{\left( 5x \right)}^{2}}{{\left( 2y \right)}^{4}}{{+}^{6}}{{C}_{5}}5x{{\left( 2y \right)}^{5}}{{+}^{6}}{{C}_{6}}{{\left( 5x \right)}^{0}}{{\left( 2y \right)}^{6}}$We know that $^{n}{{C}_{0}}=1{{,}^{n}}{{C}_{1}}=n$ and $^{n}{{C}_{n}}=1$ . Let us use these values in the above expansion.
${{\left( 5x+2y \right)}^{6}}={{\left( 5x \right)}^{6}}+6{{\left( 5x \right)}^{5}}2y{{+}^{6}}{{C}_{2}}{{\left( 5x \right)}^{4}}{{\left( 2y \right)}^{2}}{{+}^{6}}{{C}_{3}}{{\left( 5x \right)}^{3}}{{\left( 2y \right)}^{3}}{{+}^{6}}{{C}_{4}}{{\left( 5x \right)}^{2}}{{\left( 2y \right)}^{4}}{{+}^{6}}{{C}_{5}}5x{{\left( 2y \right)}^{5}}+{{\left( 2y \right)}^{6}}$
We know that $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ . Therefore, we can simplify the above expansion as follows.
${{\left( 5x+2y \right)}^{6}}={{\left( 5x \right)}^{6}}+6{{\left( 5x \right)}^{5}}2y+\dfrac{6!}{\left( 6-2 \right)!2!}{{\left( 5x \right)}^{4}}{{\left( 2y \right)}^{2}}+\dfrac{6!}{\left( 6-3 \right)!3!}{{\left( 5x \right)}^{3}}{{\left( 2y \right)}^{3}}+\dfrac{6!}{\left( 6-4 \right)!4!}{{\left( 5x \right)}^{2}}{{\left( 2y \right)}^{4}}+\dfrac{6!}{\left( 6-5 \right)!5!}\left( 5x \right){{\left( 2y \right)}^{5}}+{{\left( 2y \right)}^{6}}$
Let us simplify the above form further.
${{\left( 5x+2y \right)}^{6}}={{\left( 5x \right)}^{6}}+6{{\left( 5x \right)}^{5}}2y+\dfrac{6!}{4!2!}{{\left( 5x \right)}^{4}}{{\left( 2y \right)}^{2}}+\dfrac{6!}{3!3!}{{\left( 5x \right)}^{3}}{{\left( 2y \right)}^{3}}+\dfrac{6!}{2!4!}{{\left( 5x \right)}^{2}}{{\left( 2y \right)}^{4}}+\dfrac{6!}{1!5!}\left( 5x \right){{\left( 2y \right)}^{5}}+{{\left( 2y \right)}^{6}}$
We know that $n!=n\times \left( n-1 \right)\times \left( n-2 \right)...\times 1$ . Also we can represent any factorial as $n!=n\times \left( n-1 \right)!$ . Let us solve the above expansion.
${{\left( 5x+2y \right)}^{6}}={{\left( 5x \right)}^{6}}+6{{\left( 5x \right)}^{5}}2y+\dfrac{6\times 5\times 4!}{4!2!}{{\left( 5x \right)}^{4}}{{\left( 2y \right)}^{2}}+\dfrac{6\times 5\times 4\times 3!}{3!3!}{{\left( 5x \right)}^{3}}{{\left( 2y \right)}^{3}}+\dfrac{6\times 5\times 4!}{2!4!}{{\left( 5x \right)}^{2}}{{\left( 2y \right)}^{4}}+\dfrac{6\times 5!}{1!5!}\left( 5x \right){{\left( 2y \right)}^{5}}+{{\left( 2y \right)}^{6}}$
Let us cancel the common terms from numerator and denominator.
${{\left( 5x+2y \right)}^{6}}={{\left( 5x \right)}^{6}}+6{{\left( 5x \right)}^{5}}2y+\dfrac{6\times 5}{2!}{{\left( 5x \right)}^{4}}{{\left( 2y \right)}^{2}}+\dfrac{6\times 5\times 4}{3!}{{\left( 5x \right)}^{3}}{{\left( 2y \right)}^{3}}+\dfrac{6\times 5}{2!}{{\left( 5x \right)}^{2}}{{\left( 2y \right)}^{4}}+\dfrac{6}{1!}\left( 5x \right){{\left( 2y \right)}^{5}}+{{\left( 2y \right)}^{6}}$
Now, let us expand the factorial. We will get
${{\left( 5x+2y \right)}^{6}}={{\left( 5x \right)}^{6}}+6{{\left( 5x \right)}^{5}}2y+\dfrac{6\times 5}{2\times 1}{{\left( 5x \right)}^{4}}{{\left( 2y \right)}^{2}}+\dfrac{6\times 5\times 4}{3\times 2\times 1}{{\left( 5x \right)}^{3}}{{\left( 2y \right)}^{3}}+\dfrac{6\times 5}{2\times 1}{{\left( 5x \right)}^{2}}{{\left( 2y \right)}^{4}}+\dfrac{6}{1}\times 5x{{\left( 2y \right)}^{5}}+{{\left( 2y \right)}^{6}}$
We can now cancel the common terms from numerator and denominator.
${{\left( 5x+2y \right)}^{6}}={{\left( 5x \right)}^{6}}+6{{\left( 5x \right)}^{5}}2y+3\times 5\times {{\left( 5x \right)}^{4}}{{\left( 2y \right)}^{2}}+2\times 5\times 2{{\left( 5x \right)}^{3}}{{\left( 2y \right)}^{3}}+3\times 5\times {{\left( 5x \right)}^{2}}{{\left( 2y \right)}^{4}}+6\times 5x{{\left( 2y \right)}^{5}}+{{\left( 2y \right)}^{6}}$
On solving the above expansion, we will get
${{\left( 5x+2y \right)}^{6}}={{\left( 5x \right)}^{6}}+6{{\left( 5x \right)}^{5}}2y+15{{\left( 5x \right)}^{4}}{{\left( 2y \right)}^{2}}+20{{\left( 5x \right)}^{3}}{{\left( 2y \right)}^{3}}+15{{\left( 5x \right)}^{2}}{{\left( 2y \right)}^{4}}+30x{{\left( 2y \right)}^{5}}+{{\left( 2y \right)}^{6}}$Let us expand the powers.

& {{\left( 5x+2y \right)}^{6}}=15625{{x}^{6}}+6\times 3125{{x}^{5}}2y+15\times 625{{x}^{4}}4{{y}^{2}}+20\times 125{{x}^{3}}8{{y}^{3}}+15\times 25{{x}^{2}}16{{y}^{4}}+30x32{{y}^{5}}+64{{y}^{6}} \\\ & \Rightarrow {{\left( 5x+2y \right)}^{6}}=15625{{x}^{6}}+37500{{x}^{5}}y+37500{{x}^{4}}{{y}^{2}}+20000{{x}^{3}}{{y}^{3}}+6000{{x}^{2}}{{y}^{4}}+960x{{y}^{5}}+64{{y}^{6}} \\\ \end{aligned}$$ **Hence, the expansion of ${{\left( 5x+2y \right)}^{6}}$ is $$15625{{x}^{6}}+37500{{x}^{5}}y+37500{{x}^{4}}{{y}^{2}}+20000{{x}^{3}}{{y}^{3}}+6000{{x}^{2}}{{y}^{4}}+960x{{y}^{5}}+64{{y}^{6}}$$.** **Note:** Students must be thorough with binomial theorem as they have a chance of making mistakes when writing the formula. They must know standard combination results like $^{n}{{C}_{0}}=1{{,}^{n}}{{C}_{1}}=n$ and $^{n}{{C}_{n}}=1$ to save time. Students must be very careful when doing the mathematical calculations.