Find the coefficient of \(x^{3}\) in the expansion of \([\frac{1}{3}(2 + x)]^{6}\).
Question 1 Report
Find the coefficient of \(x^{3}\) in the expansion of \([\frac{1}{3}(2 + x)]^{6}\).
Answer Details
The binomial theorem states that the term containing \(x^r\) in the expansion of \((a+b)^n\) is given by:
$$\frac{n!}{r!(n-r)!}a^{n-r}b^{r}$$
In this case, we have \([\frac{1}{3}(2 + x)]^{6}\), so \(a = \frac{2}{3}\) and \(b = \frac{1}{3}x\). We want to find the coefficient of \(x^3\), so \(r = 3\).
Using the formula above, the term containing \(x^3\) is:
$$\frac{6!}{3!(6-3)!}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}x\right)^3 = 20\times\frac{8}{27}\times\frac{1}{27}x^3 = \frac{160}{2187}x^3$$
Therefore, the coefficient of \(x^3\) in the expansion is \(\boxed{\frac{160}{729}}\).
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