Question 1 ( 1 mark) The coefficient of \( x^{3} \) in the expansion of \( (x+2)^{5} \) is

To find the coefficient of \( x^3 \) in the expansion of \( (x+2)^{5} \), we can use the binomial theorem, which states: \[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k} \] In this case, \( n = 5 \), \( x = x \), and \( y = 2 \). We are looking for the term where \( x \) is to the power of 3, which means we need \( n-k = 3 \) or \( k = 5 - 3 = 2 \). Now, we calculate this term: \[ \text{Term} = \binom{5}{2} x^{3} (2^{2}) \] Calculating the components: 1. \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \) 2. \( 2^{2} = 4 \) Now plug in these values: \[ \text{Term} = 10 \cdot x^{3} \cdot 4 = 40 x^{3} \] The coefficient of \( x^{3} \) is therefore \( \boxed{40} \).