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Question Number 46461 by Tawa1 last updated on 26/Oct/18
Find the sum of the nth term of the series:    (1/2) + (3/4) + (7/8) + ((15)/(16)) + ...
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{7}}{\mathrm{8}}\:+\:\frac{\mathrm{15}}{\mathrm{16}}\:+\:… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Oct/18
S_n =1−(1/2)+1−(1/2^2 )+1−(1/2^3 )+1−(1/2^4 )+...  S_n =(1+1+1...)−((1/2)+(1/2^2 )+(1/2^3 )+...)  S_n =n−{(1/2)(((1−(1/2^n ))/(1−(1/2))))}  S_n =n−1+(1/2^n )
$${S}_{{n}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+… \\ $$$${S}_{{n}} =\left(\mathrm{1}+\mathrm{1}+\mathrm{1}…\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+…\right) \\ $$$${S}_{{n}} ={n}−\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)\right\} \\ $$$${S}_{{n}} ={n}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$ \\ $$
Commented by Tawa1 last updated on 26/Oct/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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