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Question Number 170580 by mnjuly1970 last updated on 27/May/22
calculate Σ_(n=1) ^∞ (( F_n )/(n.3^( n) )) = ?
$$ \\ $$$$\:\:\:{calculate} \\ $$$$\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\:\mathrm{F}_{{n}} }{{n}.\mathrm{3}^{\:{n}} }\:=\:?\:\:\: \\ $$$$ \\ $$
Answered by Mathspace last updated on 27/May/22
f_n is fefimed by f_0 =0 and?f_1 =1 and f_(n+2) =f_n +f_(n+1) ce→r^2 −r−1=0 Δ=1+4=5 ⇒r_1 =((1+(√5))/2) r_2 =((1−(√5))/2) ⇒f_n =ar_1 ^n +br_2 ^n f_0 =0=a+b f_1 =1=r_1 a+r_2 b=(r_1 −r_2 )a =(√5)a ⇒a=(1/( (√5))) and b=−(1/( (√5))) ⇒f_n =(1/( (√5)))r_1 ^n −(1/( (√5)))r_2 ^n ⇒ Σ_(n=1) ^∞ (f_n /(n3^n ))=(1/( (√5)))Σ_(n=1) ^∞ (1/n)((r_1 /3))^n −(1/( (√5)))Σ_(n=1) ^∞ (1/n)((r_2 /3))^n let s(x)=Σ_(n=1) ^∞ (x^n /n) ⇒ s(x)=−ln(1−x) ⇒ Σ_(n=1) ^∞ (f_n /(n3^n ))=−(1/( (√5)))ln(1−(r_1 /3)) +(1/( (√5)))ln(1−(r_2 /3)) =(1/( (√5)))ln(((3−r_2 )/(3−r_1 ))) =(1/( (√5)))ln(((3−((1−(√5))/2))/(3−((1+(√5))/2)))) =(1/( (√5)))ln(((5+(√5))/(5−(√5))))=(1/( (√5)))ln((((√5)+1)/( (√5)−1))) 0
$${f}_{{n}} {is}\:{fefimed}\:{by}\:{f}_{\mathrm{0}} =\mathrm{0}\:{and}?{f}_{\mathrm{1}} =\mathrm{1} \\ $$$${and}\:\:{f}_{{n}+\mathrm{2}} ={f}_{{n}} +{f}_{{n}+\mathrm{1}} \\ $$$${ce}\rightarrow{r}^{\mathrm{2}} −{r}−\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\mathrm{1}+\mathrm{4}=\mathrm{5}\:\Rightarrow{r}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${r}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow{f}_{{n}} ={ar}_{\mathrm{1}} ^{{n}} +{br}_{\mathrm{2}} ^{{n}} \\ $$$${f}_{\mathrm{0}} =\mathrm{0}={a}+{b} \\ $$$${f}_{\mathrm{1}} =\mathrm{1}={r}_{\mathrm{1}} {a}+{r}_{\mathrm{2}} {b}=\left({r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right){a} \\ $$$$=\sqrt{\mathrm{5}}{a}\:\Rightarrow{a}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:{and}\:{b}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow{f}_{{n}} \:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{r}_{\mathrm{1}} ^{{n}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{r}_{\mathrm{2}} ^{{n}} \:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{f}_{{n}} }{{n}\mathrm{3}^{{n}} }=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}}\left(\frac{{r}_{\mathrm{1}} }{\mathrm{3}}\right)^{{n}} \\ $$$$−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}}\left(\frac{{r}_{\mathrm{2}} }{\mathrm{3}}\right)^{{n}} \\ $$$${let}\:{s}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{x}^{{n}} }{{n}}\:\Rightarrow \\ $$$${s}\left({x}\right)=−{ln}\left(\mathrm{1}−{x}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{f}_{{n}} }{{n}\mathrm{3}^{{n}} }=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\left(\mathrm{1}−\frac{{r}_{\mathrm{1}} }{\mathrm{3}}\right) \\ $$$$+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\left(\mathrm{1}−\frac{{r}_{\mathrm{2}} }{\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\left(\frac{\mathrm{3}−{r}_{\mathrm{2}} }{\mathrm{3}−{r}_{\mathrm{1}} }\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\left(\frac{\mathrm{3}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}}{\mathrm{3}−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\left(\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{5}−\sqrt{\mathrm{5}}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{0} \\ $$
Commented by mnjuly1970 last updated on 28/May/22
thanks alot...
$${thanks}\:{alot}… \\ $$
Commented by Tawa11 last updated on 08/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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