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Question Number 90609 by Maclaurin Stickker last updated on 25/Apr/20

find the infinite sumΣ_(n=0) ^∞ (F_n /2^n ) where F_n =(1/(√5))(((1+(√5))/2))^(n+1) −(1/(√5))(((1−(√5))/2))^(n+1)

$${find}\:{the}\:{infinite}\:{sum}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{F}_{{n}} }{\mathrm{2}^{{n}} }\: \\ $$$${where}\:{F}_{{n}} =\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} −\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} \\ $$

Commented by abdomathmax last updated on 25/Apr/20

we have F_n =(1/(2(√5)))×(((1+(√5))^n )/2^n )−(1/(2(√5)))(((1−(√5))^n )/2^n ) ⇒ Σ_(n=0) ^∞ (F_n /2^n ) =((1+(√5))/(2(√5)))Σ_(n=0) ^∞ (((1+(√5))/4))^n −((1−(√5))/(2(√5)))Σ_(n=0) ^∞ (((1−(√5))/4))^n =((1+(√5))/(2(√5)))×(1/(1−((1+(√5))/4)))−((1−(√5))/(2(√5)))×(1/(1−((1−(√5))/4))) =(2/(√5))×((1+(√5))/(3−(√5)))−(2/(√5))×((1−(√5))/(3+(√5))) =(2/(√5))(((1+(√5))/(3−(√5)))−((1−(√5))/(3+(√5))))

$${we}\:{have}\:{F}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}×\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{{n}} }{\mathrm{2}^{{n}} }−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\frac{\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{{n}} }{\mathrm{2}^{{n}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{F}_{{n}} }{\mathrm{2}^{{n}} }\:\:=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{{n}} −\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{{n}} \\ $$$$=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}×\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}×\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}×\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}−\sqrt{\mathrm{5}}}−\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}×\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{3}+\sqrt{\mathrm{5}}} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}−\sqrt{\mathrm{5}}}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{3}+\sqrt{\mathrm{5}}}\right) \\ $$$$ \\ $$

Commented by Maclaurin Sticcker last updated on 25/Apr/20

I got the same result, 4. Thanks for your answer.

$${I}\:{got}\:{the}\:{same}\:{result},\:\mathrm{4}.\:{Thanks}\:{for} \\ $$$${your}\:{answer}. \\ $$

Commented by abdomathmax last updated on 25/Apr/20

you are welcome

$${you}\:{are}\:{welcome} \\ $$

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