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Question Number 122391 by ZiYangLee last updated on 16/Nov/20

$$\mathrm{Given}4,2,1,\frac{1}{2},\dots \mathrm{is}\mathrm{a}\mathrm{geometric}\mathrm{progression}.$$$$\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{of}(n+2)\mathrm{terms}\mathrm{of}\mathrm{this}\mathrm{progression}$$$$\mathrm{in}\mathrm{terms}\mathrm{of}n.$$

Answered by floor(10²Eta[1]) last updated on 16/Nov/20

$$\mathrm{GP}(4,2,1,\frac{1}{2},...),\mathrm{q}=\frac{1}{2}$$$$\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{{\mathrm{a}}_1(1-{\mathrm{q}}^{\mathrm{n}})}{1-\mathrm{q}}$$$${\mathrm{S}}_{\mathrm{n}+2}=\frac{{\mathrm{a}}_1(1-{\mathrm{q}}^{\mathrm{n}+2})}{1-\mathrm{q}}=\frac{4(1-{\left(\frac{1}{2}\right)}^{\mathrm{n}+2})}{1-\frac{1}{2}}$$$$=\frac{2^{\mathrm{n}+2}-1}{2^{\mathrm{n}-1}}$$

Answered by Dwaipayan Shikari last updated on 16/Nov/20

$$4+4a+4a^2+4a^3+...+4a^{n+1}$$$$4(1+a+a^2+a^3+...+a^{n+1})$$$$=4\frac{1-a^{n+2}}{1-a}(a=\frac{1}{2})$$$$=8(1-\frac{1}{2^{n+2}})=2^3-2^{1-n}$$

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