Given the geometric series: \( 3+6+12+\ldots \) to \( n \) terms. 1.3.1 Write down the general term of this series. 1.3.2 Calculate the value of \( k \) such that: \[ \sum_{p=1}^{k} \frac{3}{2}(2)^{p}=98301 \]
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The general term of the series can be expressed as \( a_n = 3 \cdot 2^{n-1} \), where \( a_1 = 3 \), \( a_2 = 6 \), \( a_3 = 12 \), and so forth, multiplying by 2 for each subsequent term. To find \( k \) such that \( \sum_{p=1}^{k} \frac{3}{2}(2)^{p} = 98301 \), we first simplify the series: \[ \sum_{p=1}^{k} \frac{3}{2}(2)^{p} = \frac{3}{2} \sum_{p=1}^{k} 2^{p} = \frac{3}{2} (2^1 + 2^2 + ... + 2^k). \] The sum of the geometric series can be calculated using the formula \( S_k = a \frac{(r^k - 1)}{r - 1} \). Here, \( a = 2 \), \( r = 2 \), and thus: \[ \sum_{p=1}^{k} 2^{p} = 2 \frac{(2^k - 1)}{1} = 2(2^k - 1). \] Substituting back into the equation gives: \[ \frac{3}{2} \cdot 2(2^k - 1) = 98301 \implies 3(2^k - 1) = 98301 \implies 2^k - 1 = \frac{98301}{3} \implies 2^k - 1 = 32767 \implies 2^k = 32768. \] Since \( 32768 = 2^{15} \), we get \( k = 15 \).
