Geometric Determine the value of P if \( \mathrm{P}=\sum_{k=1}^{13} 3^{k-5} \) rove that for any arithmetic sequence of which the first term is \( a \) and the constant lifference is \( d \), the sum to \( n \) terms can be expressed as \( S_{n}=\frac{n}{2}(2 a+(n-1) d) \).

To determine the value of \( P \), notice that \( P = \sum_{k=1}^{13} 3^{k-5} \) can be rewritten as \( P = 3^{-5} \sum_{k=1}^{13} 3^{k} \). The inner sum is a geometric series with the first term \( 3 \) (when \( k = 1 \)) and the last term \( 3^{13} \). Using the formula for the sum of a geometric series, \( S = \frac{a (r^n - 1)}{r - 1} \), where \( a = 3 \), \( r = 3 \), and \( n = 13 \), you get \( P = 3^{-5} \left( \frac{3(3^{13}-1)}{3-1} \right) = \frac{1}{243} \left( \frac{3^{14}-3}{2} \right) \). The formula for the sum of the first \( n \) terms, \( S_n = \frac{n}{2}(2a + (n-1)d) \), derives from the concept of pairing terms. If you write out the first \( n \) terms of an arithmetic sequence, you'll start with \( a \) and end with \( a + (n-1)d \). When you sum the sequence forwards and backwards, you notice that each pair sums to the same total: \( a + (a + (n-1)d) = 2a + (n-1)d \), which leads to \( n/2 \) such pairs. Thus, the formula provides a compact means to find the sum efficiently.