## Solve Lesson 5 Page 37 Math Study Topic 10 – Kite>

Topic

Consider the expansion $${\left( {x + \frac{5}{2}} \right)^{12}}$$

a) Determine the coefficient of $${x^7}$$

b) State the general term in the above binomial expansion, then state the coefficient $${a_k}$$ of $${x^k}$$ with $$0 \le k \le 12$$

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:

$${\left( {x + \frac{5}{2}} \right)^{12}} = C_{12}^0{x^{12}} + C_{12}^1{x^ {11}}{\left( {\frac{5}{2}} \right)^1} + … + C_{12}^k{x^{12 – k}}{\left( {\frac{ 5}{2}} \right)^k} + … + C_{12}^{12}{\left( {\frac{5}{2}} \right)^{12}}$$

The term containing $${x^7}$$ corresponds to $$12 – k = 7 \Rightarrow k = 5$$. Therefore the coefficient of $${x^7}$$ is

$$C_{12}^5{\left( {\frac{5}{2}} \right)^5}$$

b) The term containing $${x^k}$$ in the expansion of $${\left( {x + \frac{5}{2}} \right)^{12}}$$ is $$C_) {12}^{12 – k}{(x)^k}{\left( {\frac{5}{2}} \right)^{12 – k}}$$

Thus, the coefficient $${a_k}$$ of $${x^k}$$ with $$0 \le k \le 12$$ is $$C_{12}^{12 – k}{\left ( {\frac{5}{2}} \right)^{12 – k}}$$