Question
Evaluate the following:$\sum_\limits{\text{k}=1}^{\text{n}}(2^\text{k}+3^{\text{k}-1})$
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Answer
$\sum_\limits{\text{k}=1}^{\text{n}}(2^\text{k}+3^{\text{k}-1})$ $=(2+3^0)+(2^2+3)+(2^3+3^2)+\ \dots\ +(2^\text{n}+3^{{\text{n}-}1})$ $(2+2^2+2^3+\ \dots\ +2^{\text{n}})+(3^0+3^1+3^2+\ \dots\ +3^{\text{n}-1})$ $=\text{S}_\text{n}+\text{S}_\text{m}$ $\text{S}_\text{n}\Rightarrow\text{a}=2,\text{n}=\text{n},\text{r}=\frac{2^2}{2}=2$ $\text{S}_\text{n}=\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1}=\frac{2(2^\text{n}-1)}{2-1}=2(2^\text{n}-1)$ Also, $\text{S}_\text{m}=\text{S}_{\text{n}-1}$ $\text{a}=1,\text{r}=3,\text{n}=\text{n}-1$ $\text{S}_{\text{n}-1}=\frac{1(3^{\text{n}-1}-1)}{3-1}=\frac12(3^{\text{n}}-1)$ $\therefore\sum_\limits{\text{k}-1}^\text{n}(2^{\text{k}}+3^{\text{k}-1})=2(2^\text{n}-1)+\frac12(3^\text{n}-1)$ $=\frac12\big[2^{\text{n}+2}+3^\text{n}-4-1\big]$ $=\frac12\big[2^{\text{n}+2}+3^\text{n}-5\big]$
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