QUESTION 2 Consider the geometric series: \[ \mathrm{K}=\sum_{n=2}^{p} 1458.3^{1-n} \] \( 2.1 \quad \) Determine the smallest integral value of K . \( 2.2 \quad \) Hence, or otherwise, determine the sum of all the fractional terms in: \[ \sum_{n=2}^{\infty} 1458.3^{1-n} \]

Let \[ K=\sum_{n=2}^{p}1458\cdot 3^{\,1-n}. \] Notice that when \( n=2 \) the term is \[ 1458\cdot 3^{1-2}=1458\cdot 3^{-1}=\frac{1458}{3}=486. \] Thus the first term is \( a=486 \) and when we increase \( n \) by 1 the factor is \[ \frac{1458\cdot 3^{\,1-(n+1)}}{1458\cdot 3^{\,1-n}}=3^{-1}=\frac{1}{3}. \] So the series is geometric with common ratio \( r=\frac{1}{3} \) and with \( p-1 \) terms (since \( n \) runs from 2 to \( p \)). The formula for the sum of a geometric series is \[ S_{p-1}=a\frac{1-r^{p-1}}{1-r}. \] Substitute \( a=486 \) and \( r=\frac{1}{3} \): \[ K=486\cdot\frac{1-(1/3)^{p-1}}{1-\frac{1}{3}} =486\cdot\frac{1-(1/3)^{p-1}}{\frac{2}{3}} =486\cdot\frac{3}{2}\Bigl(1-\Bigl(\frac{1}{3}\Bigr)^{p-1}\Bigr) =729\Bigl(1-\Bigl(\frac{1}{3}\Bigr)^{p-1}\Bigr). \] Observe that \[ 729=3^6, \] so we can rewrite \[ K=3^6-\frac{3^6}{3^{p-1}} =3^6-3^{6-(p-1)} =3^6-3^{7-p}. \] It follows that for \( K \) to be an integer the term \( 3^{7-p} \) must be an integer. This is automatic provided that \( 7-p\ge0 \) (that is \( p\le7 \)). Thus, when \( p\le7 \) the sum is integral. Moreover, as \( p \) increases (from 2 up to 7) the value of \( K \) increases: - For \( p=2 \): \[ K=3^6-3^{7-2}=729-3^5=729-243=486. \] - For \( p=3 \): \[ K=729-3^{7-3}=729-3^4=729-81=648. \] - For \( p=4 \): \[ K=729-3^{7-4}=729-3^3=729-27=702. \] - For \( p=5 \): \[ K=729-3^{7-5}=729-3^2=729-9=720. \] - For \( p=6 \): \[ K=729-3^{7-6}=729-3=726. \] - For \( p=7 \): \[ K=729-3^{7-7}=729-1=728. \] Thus the smallest (least) integral value of \( K \) occurs when \( p=2 \) and is \[ 486. \] Now, for the second part we consider the full infinite series \[ \sum_{n=2}^{\infty}1458\cdot 3^{\,1-n}. \] Its sum is the limit as \( p\to\infty \). Using \[ K=729\Bigl(1-\Bigl(\frac{1}{3}\Bigr)^{p-1}\Bigr), \] taking the limit gives \[ \lim_{p\to\infty}K=729\Bigl(1-0\Bigr)=729. \] Examine the individual terms: - For \( n=2 \) to \( n=7 \) we obtain: \[ \begin{array}{ll} n=2:& 1458\cdot 3^{-1}=486,\\[1mm] n=3:& 1458\cdot 3^{-2}=162,\\[1mm] n=4:& 1458\cdot 3^{-3}=54,\\[1mm] n=5:& 1458\cdot 3^{-4}=18,\\[1mm] n=6:& 1458\cdot 3^{-5}=6,\\[1mm] n=7:& 1458\cdot 3^{-6}=2. \end{array} \] Each of these is an integer. - For \( n\ge8 \), the terms are \[ 1458\cdot 3^{\,1-n}=\frac{1458}{3^{\,n-1}}, \] which are fractional. In particular, when \( n=8 \): \[ 1458\cdot 3^{-7}=\frac{1458}{2187}=\frac{2}{3}. \] Thus the fractional part of the infinite sum comes entirely from the terms with \( n\ge8 \). The sum of the fractional terms is \[ \sum_{n=8}^{\infty}1458\cdot 3^{\,1-n}. \] Since the infinite sum of all terms is \( 729 \) and the sum of the first 6 integer terms (from \( n=2 \) to \( n=7 \)) is \[ 486+162+54+18+6+2=728, \] the sum of all fractional terms is \[ 729-728=1. \] Alternatively, note that the series of fractional terms is geometric with first term \[ a'=1458\cdot 3^{-7}=\frac{1458}{2187}=\frac{2}{3}, \] and ratio \( r=\frac{1}{3} \). Then \[ S=\frac{a'}{1-r}=\frac{\frac{2}{3}}{1-\frac{1}{3}}=\frac{\frac{2}{3}}{\frac{2}{3}}=1. \] Thus the answers are: \[ \textbf{2.1:}\quad 486, \qquad \textbf{2.2:}\quad 1. \]