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Question Number 120363 by Lordose last updated on 30/Oct/20

$$$$$$\mathrm{Determine}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}1\mathrm{st}\mathrm{nth}\mathrm{term}$$$$\mathrm{of}\mathrm{the}\mathrm{Sequence}$$$$$$$$1,4,10,22,46,....$$$$$$$$★\mathrm{Almighty}\mathrm{Formula}$$

Answered by floor(10²Eta[1]) last updated on 31/Oct/20

$$(1,4,10,22,46,...,{\mathrm{a}}_{\mathrm{n}})=({\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,...,{\mathrm{a}}_{\mathrm{n}})$$$${\mathrm{a}}_1={\mathrm{a}}_1$$$${\mathrm{a}}_2={\mathrm{a}}_1+3$$$${\mathrm{a}}_3={\mathrm{a}}_2+6$$$${\mathrm{a}}_4={\mathrm{a}}_3+12$$$$...\left(\mathrm{sum}\mathrm{all}\mathrm{together}\right)$$$${\mathrm{S}}_{\mathrm{n}}={\mathrm{a}}_1+{\mathrm{S}}_{\mathrm{n}-1}+(3+6+12+...+{\mathrm{b}}_{\mathrm{n}-1})$$$${\mathrm{S}}_{\mathrm{n}}={\mathrm{a}}_1+{\mathrm{S}}_{\mathrm{n}}-{\mathrm{a}}_{\mathrm{n}}+(3+6+...+{\mathrm{b}}_{\mathrm{n}-1})$$$${\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_1+(3+6+12+...+{\mathrm{b}}_{\mathrm{n}-1})$$$${\mathrm{a}}_{\mathrm{n}}=1+(3+6+12+...+{\mathrm{b}}_{\mathrm{n}-1})\left[\mathrm{geometric}\mathrm{progression}\right]$$$${\mathrm{a}}_{\mathrm{n}}=1+3(2^{\mathrm{n}-1}-1)$$$${\mathrm{a}}_{\mathrm{n}}=3.2^{\mathrm{n}-1}-2$$$${\mathrm{S}}_{\mathrm{n}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}(3.2^{\mathrm{k}-1}-2)=3\left(\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{2}^{\mathrm{k}-1}\right)-2\mathrm{n}$$$$\Rightarrow {\mathrm{S}}_{\mathrm{n}}=3.2^{\mathrm{n}}-3-2\mathrm{n}$$$$$$

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