Question Number 122023 by Dwaipayan Shikari last updated on 13/Nov/20
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {F}_{{n}} }{\mathrm{7}^{{n}} }\:\:\:\:\:\:\left({F}_{{n}} \:{denotes}\:{Fibbonocci}\:{sequence}\right) \\ $$
Answered by mr W last updated on 13/Nov/20
![F_n =(1/( (√5)))(φ^n −ψ^n ) with φ=((1+(√5))/2) ψ=((1−(√5))/2) S=Σ_(n=0) ^∞ (((−1)^n F_n )/7^n ) =(1/( (√5)))Σ_(n=0) ^∞ [(−(φ/7))^n −(−(ψ/7))^n ] =(1/( (√5)))((1/(1+(φ/7)))−(1/(1+(ψ/7)))) =(1/( (√5)))((1/(1+((1+(√5))/(14))))−(1/(1−(((√5)−1)/(14))))) =((14)/( (√5)))((1/(15+(√5)))−(1/(15−(√5)))) =−(7/( 55))](Q122029.png)
$${F}_{{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\phi^{{n}} −\psi^{{n}} \right)\:{with} \\ $$$$\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\psi=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {F}_{{n}} }{\mathrm{7}^{{n}} } \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(−\frac{\phi}{\mathrm{7}}\right)^{{n}} −\left(−\frac{\psi}{\mathrm{7}}\right)^{{n}} \right] \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\phi}{\mathrm{7}}}−\frac{\mathrm{1}}{\mathrm{1}+\frac{\psi}{\mathrm{7}}}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{14}}}−\frac{\mathrm{1}}{\mathrm{1}−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{14}}}\right) \\ $$$$=\frac{\mathrm{14}}{\:\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}}{\mathrm{15}+\sqrt{\mathrm{5}}}−\frac{\mathrm{1}}{\mathrm{15}−\sqrt{\mathrm{5}}}\right) \\ $$$$=−\frac{\mathrm{7}}{\:\mathrm{55}} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Nov/20
$${Typo}\:\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{14}}}=\frac{\mathrm{14}}{\mathrm{15}−\sqrt{\mathrm{5}}} \\ $$
Commented by Dwaipayan Shikari last updated on 14/Nov/20
$${Thanking}\:{you} \\ $$