$$\sum_{k=1}^{11}{2+3^k}$$
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MathBot Answer:
\[\sum_{k=1}^{11} \left(3^{k} + 2\right) = 265741\]
$$\begin{aligned}\sum_{k=1}^{11} \left(3^{k} + 2\right) &= \sum_{k=1}^{11} 2 + \sum_{k=1}^{11} 3^{k} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} \left(f{\left(n \right)} + g{\left(n \right)}\right) = \sum_{n=k}^{m} f{\left(n \right)} + \sum_{n=k}^{m} g{\left(n \right)}}\\&= \sum_{k=1}^{11} 3^{k} + 11 \cdot 2 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=1}^{m} C = mC}\\&= \sum_{k=1}^{11} 3^{k} + 22\\&= \sum_{k=0}^{10} 3^{k + 1} + 22 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k + p}^{m + p} f{\left(n - p \right)} = \sum_{n=k}^{m} f{\left(n \right)}}\\&= \sum_{k=0}^{10} 3 \cdot 3^{k} + 22\\&= 3 \sum_{k=0}^{10} 3^{k} + 22 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} C f{\left(n \right)} = C \sum_{n=k}^{m} f{\left(n \right)}}\\&= \frac{3 \cdot \left(1 - 3^{1 + 10}\right)}{\left(1 - 3\right)} + 22 \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} a^n = \frac{1-a^{m+1}}{1-a}}\\&= 22 + 265719\\&= 265741\end{aligned}$$