Question Number 224042 by Tawa11 last updated on 15/Aug/25
Answered by mr W last updated on 16/Aug/25
![let a_n =b_n +k b_(n+2) +k=2(b_(n+1) +k)+b_n +k−1 b_(n+2) =2b_(n+1) +b_n +2k−1 let 2k−1=0, i.e. k=(1/2), then b_(n+2) =2b_(n+1) +b_n , i.e. b_(n+2) −2b_(n+1) −b_n =0 the characteristic equation is r^2 −2r−1=0 ⇒r=1±(√2) ⇒b_n =A(1+(√2))^n +B(1−(√2))^n ⇒a_n =A(1+(√2))^n +B(1−(√2))^n +(1/2) a_0 =A+B+(1/2)=1 ...(i) a_1 =A(1+(√2))+B(1−(√2))+(1/2)=1 ...(ii) ⇒A=B=(1/4) ⇒a_n =(1/4)[(1+(√2))^n +(1−(√2))^n ]+(1/2) (a_k /3^k )=(1/4)[(((1+(√2))/3))^k +(((1−(√2))/3))^k ]+(1/2)((1/3))^k ∣((1+(√2))/3)∣<1 ∣((1−(√2))/3)∣<1 ∣(1/3)∣<1 for ∣q∣<1, we have Σ_(k=0) ^∞ q^k =(1/(1−q)), hence S=Σ_(k=0) ^∞ (a_k /3^k )=(1/4)[Σ_(k=0) ^∞ (((1+(√2))/3))^k +Σ_(k=0) ^∞ (((1−(√2))/3))^k ]+(1/2)Σ_(k=0) ^∞ (1/3^k ) =(1/4)[(1/(1−((1+(√2))/3)))+(1/(1−((1−(√2))/3)))]+(1/2)×(1/(1−(1/3))) =(1/4)[((3(2+(√2)))/2)+((3(2−(√2)))/2)]+(1/2)×(3/2) =(9/4) ✓](https://www.tinkutara.com/question/Q224045.png)
$${let}\:{a}_{{n}} ={b}_{{n}} +{k} \\ $$$${b}_{{n}+\mathrm{2}} +{k}=\mathrm{2}\left({b}_{{n}+\mathrm{1}} +{k}\right)+{b}_{{n}} +{k}−\mathrm{1} \\ $$$${b}_{{n}+\mathrm{2}} =\mathrm{2}{b}_{{n}+\mathrm{1}} +{b}_{{n}} +\mathrm{2}{k}−\mathrm{1} \\ $$$${let}\:\mathrm{2}{k}−\mathrm{1}=\mathrm{0},\:{i}.{e}.\:{k}=\frac{\mathrm{1}}{\mathrm{2}},\:{then} \\ $$$${b}_{{n}+\mathrm{2}} =\mathrm{2}{b}_{{n}+\mathrm{1}} +{b}_{{n}} ,\:{i}.{e}. \\ $$$${b}_{{n}+\mathrm{2}} −\mathrm{2}{b}_{{n}+\mathrm{1}} −{b}_{{n}} =\mathrm{0} \\ $$$${the}\:{characteristic}\:{equation}\:{is} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{r}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{r}=\mathrm{1}\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{b}_{{n}} ={A}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{{n}} +{B}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)^{{n}} \\ $$$$\Rightarrow{a}_{{n}} ={A}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{{n}} +{B}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)^{{n}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}_{\mathrm{0}} ={A}+{B}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1}\:\:\:…\left({i}\right) \\ $$$${a}_{\mathrm{1}} ={A}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+{B}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow{A}={B}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{a}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{{n}} +\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)^{{n}} \right]+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\frac{{a}_{{k}} }{\mathrm{3}^{{k}} }=\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{3}}\right)^{{k}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{3}}\right)^{{k}} \right]+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{k}} \\ $$$$\mid\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{3}}\mid<\mathrm{1} \\ $$$$\mid\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{3}}\mid<\mathrm{1} \\ $$$$\mid\frac{\mathrm{1}}{\mathrm{3}}\mid<\mathrm{1} \\ $$$${for}\:\mid{q}\mid<\mathrm{1},\:{we}\:{have}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{q}^{{k}} =\frac{\mathrm{1}}{\mathrm{1}−{q}},\:{hence} \\ $$$${S}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{a}_{{k}} }{\mathrm{3}^{{k}} }=\frac{\mathrm{1}}{\mathrm{4}}\left[\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{3}}\right)^{{k}} +\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{3}}\right)^{{k}} \right]+\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{3}^{{k}} } \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{3}}}\right]+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}}+\frac{\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\right]+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{9}}{\mathrm{4}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 16/Aug/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Commented by necx122 last updated on 16/Aug/25
$${This}\:{solution}\:{is}\:{beautiful}. \\ $$