Question

How do you expand ${{\left( 2x+5 \right)}^{7}}$?

Answer 1

In the above question, we have been given a binomial which is raised to a power. For expanding it, we have to use the binomial theorem. The binomial theorem states that the expansion of the binomial ${{\left( a+b \right)}^{n}}$ is given by $\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}}$. In the case of the above question, we have $a=2x$, $b=5$ and $n=7$. Therefore, on putting these into the summation $\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}}$ and on substituting the values of r from zero to $7$, we will obtain the expansion of the given expression in the form of a series.

Complete step by step answer:
Let us consider the expression given in the above question as
$\Rightarrow E={{\left( 2x+5 \right)}^{7}}$
As we can observe, the above expression is a binomial raised to the power of $7$. Therefore, we can use the binomial theorem to expand it which is be given by
$\Rightarrow {{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}}$
On comparing the above equation with the given expression, we get $a=2x$, $b=5$ and $n=7$. Therefore, on substituting these in the above equation, we get
$\Rightarrow {{\left( 2x+5 \right)}^{7}}=\sum\limits_{r=0}^{7}{^{7}{{C}_{r}}{{\left( 2x \right)}^{7-r}}{{\left( 5 \right)}^{r}}}$
On substituting the values of r from zero to seven, we will obtain

& \Rightarrow {{\left( 2x+5 \right)}^{7}}{{=}^{7}}{{C}_{0}}{{\left( 2x \right)}^{7-0}}{{\left( 5 \right)}^{0}}{{+}^{7}}{{C}_{1}}{{\left( 2x \right)}^{7-1}}{{\left( 5 \right)}^{1}}{{+}^{7}}{{C}_{2}}{{\left( 2x \right)}^{7-2}}{{\left( 5 \right)}^{2}}{{+}^{7}}{{C}_{3}}{{\left( 2x \right)}^{7-3}}{{\left( 5 \right)}^{3}}{{+}^{7}}{{C}_{4}}{{\left( 2x \right)}^{7-4}}{{\left( 5 \right)}^{4}} \\\ & {{+}^{7}}{{C}_{5}}{{\left( 2x \right)}^{7-5}}{{\left( 5 \right)}^{5}}{{+}^{7}}{{C}_{6}}{{\left( 2x \right)}^{7-6}}{{\left( 5 \right)}^{6}}{{+}^{7}}{{C}_{7}}{{\left( 2x \right)}^{7-7}}{{\left( 5 \right)}^{7}} \\\ \end{aligned}$$$$\begin{aligned} & \Rightarrow {{\left( 2x+5 \right)}^{7}}={{\left( 2x \right)}^{7}}+7{{\left( 2x \right)}^{6}}{{\left( 5 \right)}^{1}}+21{{\left( 2x \right)}^{5}}{{\left( 5 \right)}^{2}}+35{{\left( 2x \right)}^{4}}{{\left( 5 \right)}^{3}}+35{{\left( 2x \right)}^{3}}{{\left( 5 \right)}^{4}}+21{{\left( 2x \right)}^{2}}{{\left( 5 \right)}^{5}} \\\ & +7{{\left( 2x \right)}^{1}}{{\left( 5 \right)}^{6}}+{{\left( 5 \right)}^{7}} \\\ \end{aligned}$$ **Note:** There are huge chances of error while substituting the values of r from zero to n. The trick to avoid this is that the sum of the exponents over a and b in the summation ${{\left( a+b \right)}^{n}}$ must be equal to n in each term of the series. Further, we may also take out ${{2}^{7}}$ common from the given expression ${{\left( 2x+5 \right)}^{7}}$ before applying the binomial theorem for expanding it.