## Solve Lesson 6 Page 37 Math Study Topic 10 – Kite>

Topic

Consider the expansion $${\left( {\frac{x}{2} + \frac{1}{5}} \right)^{21}}$$

a) Determine the coefficient of $${x^{10}}$$

b) State the general term in the above binomial expansion, then state the coefficient $${a_k}$$ of $${x^k}$$ with $$0 \le k \le 21$$

Solution method – See details Newton’s binomial formula: $${(a + b)^n} = C_n^0{a^n} + C_n^1{a^{n – 1}}b + … + C_n^{n – 1} a{b^{n – 1}} + C_n^n{b^n}$$

Detailed explanation

a) According to Newton’s binomial formula, we have:

$${\left( {\frac{x}{2} + \frac{1}{5}} \right)^{21}} = C_{21}^0{\left( {\frac{x} {2}} \right)^{12}} + C_{21}^1{\left( {\frac{x}{2}} \right)^{20}}{\left( {\frac{1 }{5}} \right)^1} + … + C_{21}^k{\left( {\frac{x}{2}} \right)^{21 – k}}{\left( {\ frac{1}{5}} \right)^k} + … + C_{21}^{21}{\left( {\frac{1}{5}} \right)^{21}}$$

The term containing $${x^{10}}$$ corresponds to $$21 – k = 10 \Rightarrow k = 11$$. Therefore the coefficient of $${x^{10}}$$ is

$$C_{21}^{11}{\left( {\frac{1}{2}} \right)^{10}}{\left( {\frac{1}{5}} \right)^ {11}}$$

b) Term containing $${x^k}$$ in the expansion of $${\left( {\frac{x}{2} + \frac{1}{5}} \right)^{21} }$$ is $$C_{21}^{21 – k}{\left( {\frac{x}{2}} \right)^k}{\left( {\frac{1}{5}} \right)^{21 – k}}$$

Thus, the coefficient $${a_k}$$ of $${x^k}$$ with $$0 \le k \le 21$$ is $$C_{21}^{21 – k}{\left ( {\frac{1}{2}} \right)^k}{\left( {\frac{1}{5}} \right)^{21 – k}}$$