Answer
**2.1:** The smallest integral value of \( K \) is **486**.
**2.2:** The sum of all the fractional terms in the infinite series is **1**.
Solution
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:
\[
\textbf{2.1:}\quad 486, \qquad \textbf{2.2:}\quad 1.
\]
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