Question

How do you find the coefficient of $x^{6}$ in the expansion of ${{\left( 2x+3 \right)}^{10}}$ ?

Answer 1

In order to do this question, first you need to write the above term in the form of binomial expansion that is ${{\left( ax+b \right)}^{y}}=\sum\limits_{r=0}^{y}{{}^{y}{{C}_{r}}{{a}^{y-r}}{{b}^{r}}{{x}^{y-k}}}$. Then you can find the value of r for which you get the required power of x. In this you need to substitute a with 2, b with 3, y with 10. Then you get the final answer for the coefficient of the required term.

Complete step by step solution:
In order to do this question, first you need to write the above term in the form of binomial expansion that is ${{\left( ax+b \right)}^{y}}=\sum\limits_{r=0}^{y}{{}^{y}{{C}_{r}}{{a}^{y-r}}{{b}^{r}}{{x}^{y-k}}}$.
$\Rightarrow {{\left( ax+b \right)}^{y}}=\sum\limits_{r=0}^{y}{{}^{y}{{C}_{r}}{{a}^{y-r}}{{b}^{r}}{{x}^{y-k}}}$.
. In this you need to substitute a with 2, b with 3, y with 10. Then, we get
$\Rightarrow {{\left( 2x+3 \right)}^{10}}=\sum\limits_{r=0}^{10}{{}^{10}{{C}_{r}}{{2}^{10-r}}{{3}^{r}}{{x}^{10-k}}}$.
Here, to get the required coefficient that is , we need to find the value of r. Therefore, we get the value of r as 4. Therefore, the coefficient is:
$\Rightarrow {}^{10}{{C}_{4}}{{2}^{6}}{{3}^{4}}= \dfrac{10!}{6!4!}\times 64\times 81$

$\Rightarrow {}^{10}{{C}_{4}}{{2}^{6}}{{3}^{4}}=210\times 64\times 81$
$\Rightarrow {}^{10}{{C}_{4}}{{2}^{6}}{{3}^{4}}=1088640$

Therefore, we find the coefficient of $x^{6}$ in the expansion of ${{\left( 2x+3 \right)}^{10}}$ is 1088640.

Note: In order to do this question, you need to know how to write the binomial expansion. Otherwise, you will not be able to do this question. You can also find the value of ${}^{10}{{C}_{4}}$ from the pascal's triangle. You get the answer at row 10 and column 4. Also, you will have to be careful in substitutions, otherwise, you will get the wrong answer in the end.