4. Find the coefficient of \( a^{2} \) in expansion of \( (2 a+1)^{5} \)

To find the coefficient of \( a^{2} \) in the expansion of \( (2a + 1)^{5} \), we can use the binomial theorem, which states that: \[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{k} y^{n-k} \] In our case, let \( x = 2a \), \( y = 1 \), and \( n = 5 \). We need to find the term where the power of \( a \) is 2. This corresponds to the value of \( k \) in \( x^{k} = (2a)^{k} \) being equal to 2. Therefore, \( k = 2 \) and \( n-k = 5-2 = 3 \). The term in the expansion is given by: \[ \binom{5}{2} (2a)^{2} (1)^{3} \] Now, compute each part: 1. The binomial coefficient \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \). 2. Calculate \( (2a)^{2} = 4a^{2} \). Putting it all together, we have: \[ \binom{5}{2} (2a)^{2} = 10 \times 4a^{2} = 40a^{2} \] Thus, the coefficient of \( a^{2} \) is \( 40 \).