1) \( \frac{\text { Simplify }}{2 x+3} \) \( \frac{2^{2 x+3}-5.2^{2 x+1}}{4^{x+2}} \) \( \frac{1,2\left(a b^{2} c\right)^{3}}{a^{-1} b^{4} c^{3}} \times \frac{\left(a^{-2} b^{3} c^{5}\right)^{0}}{a^{-1} b^{4} c^{2}} \) 2) Arithmetic Sequence \( \& \) deries 2.1 Given \( S_{n}=2970 \), and the Sequence \( 12 ; 18 ; 24 \ldots \). Find number of terms ( \( n \) )

To tackle the first expression, simplify \( \frac{2^{2x+3} - 5 \cdot 2^{2x+1}}{4^{x+2}} \): Start by rewriting \( 4^{x+2} \) as \( 2^{4x+4} \). Then, notice \( 2^{2x+3} \) can be factored as \( 2^{2x+1}(2^2 - 5) = 2^{2x+1}(-1) \). Therefore, you'll end up with \( \frac{-2^{2x+1}}{2^{4x+4}} = \frac{-1}{2^{2x + 3}} \). Now, for the second part regarding \( S_n = 2970 \) in the arithmetic sequence \( 12, 18, 24, \ldots \): The common difference here is 6. Use the formula for the sum of the first n terms, \( S_n = \frac{n}{2} (a + l) \), where \( a \) is the first term and \( l \) is the last term. Setting up the equation gives you \( S_n = \frac{n}{2} (12 + (12 + (n-1) \cdot 6)) \). Solving it will yield the number of terms \( n = 30 \).