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Given-4-2-1-1-2-is-a-geometric-progression-Find-the-sum-of-n-2-terms-of-this-progression-in-terms-of-n-




Question Number 122391 by ZiYangLee last updated on 16/Nov/20
Given 4,2,1,(1/2),… is a geometric progression.  Find the sum of (n+2)terms of this progression  in terms of n.
$$\mathrm{Given}\:\mathrm{4},\mathrm{2},\mathrm{1},\frac{\mathrm{1}}{\mathrm{2}},\ldots\:\mathrm{is}\:\mathrm{a}\:\mathrm{geometric}\:\mathrm{progression}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\left({n}+\mathrm{2}\right)\mathrm{terms}\:\mathrm{of}\:\mathrm{this}\:\mathrm{progression} \\ $$$$\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n}. \\ $$
Answered by floor(10²Eta[1]) last updated on 16/Nov/20
GP(4, 2, 1, (1/2), ...), q=(1/2)  ⇒S_n =((a_1 (1−q^n ))/(1−q))  S_(n+2) =((a_1 (1−q^(n+2) ))/(1−q))=((4(1−((1/2))^(n+2) ))/(1−(1/2)))  =((2^(n+2) −1)/2^(n−1) )
$$\mathrm{GP}\left(\mathrm{4},\:\mathrm{2},\:\mathrm{1},\:\frac{\mathrm{1}}{\mathrm{2}},\:…\right),\:\mathrm{q}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{S}_{\mathrm{n}} =\frac{\mathrm{a}_{\mathrm{1}} \left(\mathrm{1}−\mathrm{q}^{\mathrm{n}} \right)}{\mathrm{1}−\mathrm{q}} \\ $$$$\mathrm{S}_{\mathrm{n}+\mathrm{2}} =\frac{\mathrm{a}_{\mathrm{1}} \left(\mathrm{1}−\mathrm{q}^{\mathrm{n}+\mathrm{2}} \right)}{\mathrm{1}−\mathrm{q}}=\frac{\mathrm{4}\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{2}} \right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{2}^{\mathrm{n}+\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} } \\ $$
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((1−a^(n+2) )/(1−a))     (a=(1/2))  =8(1−(1/2^(n+2) ))=2^3 −2^(1−n)
$$\mathrm{4}+\mathrm{4}{a}+\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{3}} +…+\mathrm{4}{a}^{{n}+\mathrm{1}} \\ $$$$\mathrm{4}\left(\mathrm{1}+{a}+{a}^{\mathrm{2}} +{a}^{\mathrm{3}} +…+{a}^{{n}+\mathrm{1}} \right) \\ $$$$=\mathrm{4}\frac{\mathrm{1}−{a}^{{n}+\mathrm{2}} }{\mathrm{1}−{a}}\:\:\:\:\:\left({a}=\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{2}} }\right)=\mathrm{2}^{\mathrm{3}} −\mathrm{2}^{\mathrm{1}−{n}} \\ $$

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