**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 \({a_0} + {a_1} + {a_2} + {a_3} + {a_4} = \frac{{625}}{{81}}\)