Question
Evaluate the following:$\sum_\limits{\text{n}=2}^{10}(4^\text{n})$

Answer

$\sum_\limits{\text{n}=2}^{10}(4^\text{n})$ $=4^2+4^3+4^4+\ \dots\ +4^{10}$ $\text{a}=4^2,\text{r}=\frac{4^3}{4}=4,\text{n}=9$ $\text{S}_{10}=\frac{\text{a}(\text{r}^9-1)}{1-\text{r}}$ $=\frac{4^2(4^9-1)}{4-1}$ $=\frac13\big[4^{11}-16\big]$ $=\frac{16}{3}\big[4^{9}-1\big]$

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