Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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)

$$\mathrm{4}+\mathrm{4}{a}+\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{3}} +...+\mathrm{4}{a}^{{n}+\mathrm{1}} \\ $$$$\mathrm{4}\left(\mathrm{1}+{a}+{a}^{\mathrm{2}} +{a}^{\mathrm{3}} +...+{a}^{{n}+\mathrm{1}} \right) \\ $$$$=\mathrm{4}\frac{\mathrm{1}−{a}^{{n}+\mathrm{2}} }{\mathrm{1}−{a}}\:\:\:\:\:\left({a}=\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\mathrm{8}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{2}} }\right)=\mathrm{2}^{\mathrm{3}} −\mathrm{2}^{\mathrm{1}−{n}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com