 ## Solution 35 Page 16 Math Workbook 10 – Kite>

Topic

For $${\left( {2x – \frac{1}{3}} \right)^4} = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3 } + {a_4}{x^4}$$. Calculate:

a) $${a_2}$$

b) $${a_0} + {a_1} + {a_2} + {a_3} + {a_4}$$

Solution method – See details Step 1: Apply the expansion formula: $${(a – b)^4} = {a^4} – 4{a^3}b + 6{a^2}{b^2} – 4a{ b^3} + {b^4}$$ with $$a = 2x,b = \frac{1}{3}$$

Step 2: Replace x = 1 into the expansion in the hypothesis to sum the coefficients of the expansion

Detailed explanation

a) We have:  We see that $${a_2}$$ is the coefficient of $${x^2}$$

The term containing $${x^2}$$ in the expression expansion $${\left( {2x – \frac{1}{3}} \right)^4}$$ is $$\frac{8 }{3}{x^2}$$

The coefficient of in the expression expansion $${\left( {2x – \frac{1}{3}} \right)^4}$$ is $$\frac{8}{3}$$

So $${a_2} = \frac{8}{3}$$

b) We have $${\left( {2x – \frac{1}{3}} \right)^4} = {a_0} + {a_1}x + {a_2}{x^2} + {a_3} {x^3} + {a_4}{x^4}$$

Select x = 1, we get:  So $${a_0} + {a_1} + {a_2} + {a_3} + {a_4} = \frac{{625}}{{81}}$$