Evaluate $\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)} $

We need to evaluate the sum $\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)}$.
Using the linearity property of summation,we have:
$\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right) = } \sum\limits_{k = 1}^{11} {2 + } \sum\limits_{k = 1}^{11} {{3^k}} = 2 \times 11 + \sum\limits_{k = 1}^{11} {{3^k}} = 22 + \sum\limits_{k = 1}^{11} {{3^k}} \quad \dots (1)$
The sum $\sum\limits_{k = 1}^{11} {{3^k}}$ is a geometric series with first term $a = 3$,common ratio $r = 3$,and number of terms $n = 11$.
The sum of a geometric series is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.
Substituting the values,we get:
$S_{11} = \frac{3(3^{11} - 1)}{3 - 1} = \frac{3}{2}(3^{11} - 1)$.
Substituting this back into equation $(1)$:
$\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right) = 22 + \frac{3}{2}(3^{11} - 1)}$.