Find the sum to n terms of the series 6+9+21+69+261+......

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$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{following} \mathrm{series}. \\ 6+9+21+69+261+\cdots \cdots \\ \mathrm{Suppose} \mathrm{that}, \\ {\mathrm{S}}_{\mathrm{n}}=6+9+21+69+261+\cdots \cdots \\ \mathrm{Rewrite} \mathrm{this} \mathrm{sum} \mathrm{one} \mathrm{over} \mathrm{the} \mathrm{another} \mathrm{as} \mathrm{follows}: \\ {\mathrm{S}}_{\mathrm{n}}=6+9+21+69+261+\cdots \cdots +{\mathrm{a}}_{\mathrm{n}-1}+{\mathrm{a}}_{\mathrm{n}} \\ {\mathrm{S}}_{\mathrm{n}}= 6+9 +21+ 69+\cdots \cdots +{\mathrm{a}}_{\mathrm{n}-2}+{\mathrm{a}}_{\mathrm{n}-1}+{\mathrm{a}}_{\mathrm{n}} \\ \mathrm{Subtract} \mathrm{the} \mathrm{above} \mathrm{to} \mathrm{get}, \\ 0=6+\left[\left(9-6\right)+\left(21-9\right)+\left(69-21\right)+\left(261-69\right)+\cdots \cdots \left(\mathrm{n}-1\right) \mathrm{terms}\right]-{\mathrm{a}}_{\mathrm{n}} \\ 0=6+\left[3+12+48+192+\cdots \cdots \left(\mathrm{n}-1\right) \mathrm{terms}\right]-{\mathrm{a}}_{\mathrm{n}} \\ \mathrm{This} \mathrm{further} \mathrm{implies} \mathrm{that}, \\ {\mathrm{a}}_{\mathrm{n}}=6+\left[3+12+48+192+\cdots \cdots \left(\mathrm{n}-1\right) \mathrm{terms}\right] \\ \mathrm{The} \mathrm{series} \mathrm{within} \mathrm{the} \mathrm{bracket} \mathrm{is} \mathrm{a} \mathrm{geometric} \mathrm{progression} \mathrm{with} \mathrm{a}=3 \mathrm{and} \mathrm{r}=4 \\ \mathrm{So} \mathrm{its} \mathrm{sum} \mathrm{will} \mathrm{be} \\ 3+12+48+192+\cdots \cdots \left(\mathrm{n}-1\right) \mathrm{terms}=\frac{3\left(4^{\mathrm{n}-1}-1\right) }{4-1} \\ =\frac{3\left(4^{\mathrm{n}-1}-1\right) }{3} \\ =4^{\mathrm{n}-1}-1 \\ \mathrm{This} \mathrm{gives}, \\ {\mathrm{a}}_{\mathrm{n}}=6+4^{\mathrm{n}-1}-1 \\ {\mathrm{a}}_{\mathrm{n}}=5+4^{\mathrm{n}-1} \\ \mathrm{So} \mathrm{the} \mathrm{required} \mathrm{sum} \mathrm{is} \mathrm{given} \mathrm{by}, \\ {\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}} \left(5+4^{\mathrm{k}-1}\right) \\ =\sum _{\mathrm{k}=1}^{\mathrm{n}} \left(5\right)+\sum _{\mathrm{k}=1}^{\mathrm{n}} \left(4^{\mathrm{k}-1}\right) \\ = \left(5+5+5+\cdots \cdots \mathrm{n} \mathrm{terms} \right)+ \left(4^{1-1}+4^{2-1}+4^{3-1}+\cdots \cdots \mathrm{n} \mathrm{terms}\right) \\ = 5\mathrm{n}+ \left(1+4+4^2+\cdots \cdots \mathrm{n} \mathrm{terms}\right) \\ = 5\mathrm{n}+ \frac{\left(1\right)\left(4^{\mathrm{n}}-1\right) }{4-1} \\ =\mathbf{ }\mathbf{5}\mathbf{n}\mathbf{+}\mathbf{ }\frac{\left({\mathbf{4}}^{\mathbf{n}}\mathbf{-}\mathbf{1}\right)\mathbf{ }}{\mathbf{3}} \end{gathered}$$

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