Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 121949 by liberty last updated on 12/Nov/20

Evaluate the sum (2/(3+1)) + (2^2 /(3^2 +1)) + (2^3 /(3^4 +1))+ ...+ (2^(n+1) /(3^2^n +1)) .

$$\:\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{sum}\:\:\frac{\mathrm{2}}{\mathrm{3}+\mathrm{1}}\:+\:\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} +\mathrm{1}}\:+\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{4}} +\mathrm{1}}+\:...+\:\frac{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }{\mathrm{3}^{\mathrm{2}^{\mathrm{n}} } \:+\mathrm{1}}\:. \\ $$

Answered by bobhans last updated on 12/Nov/20

consider (1/(a+1))= ((a−1)/(a^2 +1))=(a/(a^2 −1))−(1/(a^2 −1)) =−(1/(a^2 −1)) + (1/2)((1/(a+1))+(1/(a−1))) which implies (1/2).(1/(a+1))=(1/(2(a−1)))−(1/(a^2 −1)) substitute identity with a=3^2^k we obtain (1/(2(3^2^k +1))) = (1/(2(3^2^k −1)))−(1/(3^2^(k+1) −1)) multiplying this by 2^(k+2) , we get the relation (2^(k+1) /(3^2^k +1)) = (2^(k+1) /(3^2^k −1))−(2^(k+2) /(3^2^(k+1) −1)) Thus (2/(3+1))+(2^2 /(3^2 +1))+(2^3 /(3^4 +1))+...+(2^(n+1) /(3^2^n +1)) = Σ_(k= 0) ^n ((2^(k+1) /(3^2^k −1)) −(2^(k+2) /(3^2^(k+1) −1)) ) = 1−(2^(n+2) /(3^2^(n+1) −1)).

$${consider}\:\frac{\mathrm{1}}{{a}+\mathrm{1}}=\:\frac{{a}−\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}=\frac{{a}}{{a}^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:=−\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{a}+\mathrm{1}}+\frac{\mathrm{1}}{{a}−\mathrm{1}}\right) \\ $$$${which}\:{implies}\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{{a}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}\left({a}−\mathrm{1}\right)}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$${substitute}\:{identity}\:{with}\:{a}=\mathrm{3}^{\mathrm{2}^{{k}} } \\ $$$${we}\:{obtain}\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{3}^{\mathrm{2}^{{k}} } \:+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{3}^{\mathrm{2}^{{k}} } −\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}^{{k}+\mathrm{1}} } −\mathrm{1}} \\ $$$${multiplying}\:{this}\:{by}\:\mathrm{2}^{{k}+\mathrm{2}} \:,\:{we}\:{get}\: \\ $$$${the}\:{relation}\:\frac{\mathrm{2}^{{k}+\mathrm{1}} }{\mathrm{3}^{\mathrm{2}^{{k}} } +\mathrm{1}}\:=\:\frac{\mathrm{2}^{{k}+\mathrm{1}} }{\mathrm{3}^{\mathrm{2}^{{k}} } −\mathrm{1}}−\frac{\mathrm{2}^{{k}+\mathrm{2}} }{\mathrm{3}^{\mathrm{2}^{{k}+\mathrm{1}} } −\mathrm{1}} \\ $$$${Thus}\: \\ $$$$\frac{\mathrm{2}}{\mathrm{3}+\mathrm{1}}+\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{4}} +\mathrm{1}}+...+\frac{\mathrm{2}^{{n}+\mathrm{1}} }{\mathrm{3}^{\mathrm{2}^{{n}} } +\mathrm{1}}\:= \\ $$$$\underset{{k}=\:\mathrm{0}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{2}^{{k}+\mathrm{1}} }{\mathrm{3}^{\mathrm{2}^{{k}} } −\mathrm{1}}\:−\frac{\mathrm{2}^{{k}+\mathrm{2}} }{\mathrm{3}^{\mathrm{2}^{{k}+\mathrm{1}} } −\mathrm{1}}\:\right) \\ $$$$=\:\mathrm{1}−\frac{\mathrm{2}^{{n}+\mathrm{2}} }{\mathrm{3}^{\mathrm{2}^{{n}+\mathrm{1}} } −\mathrm{1}}.\: \\ $$

Commented by liberty last updated on 12/Nov/20

good....

$$\mathrm{good}.... \\ $$

Commented by sewak last updated on 17/Nov/20

Quite good

$$\mathrm{Quite}\:\mathrm{good} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com