Question Number 176746 by mr W last updated on 26/Sep/22
$${a}_{{n}+\mathrm{2}} −\mathrm{5}{a}_{{n}+\mathrm{1}} +\mathrm{6}{a}_{{n}} =\mathrm{3}{n}+\mathrm{5}^{{n}} \\ $$$${a}_{\mathrm{1}} =\mathrm{1},\:{a}_{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:{a}_{{n}} \\ $$
Answered by FongXD last updated on 26/Sep/22
Commented by Tawa11 last updated on 26/Sep/22
$$\mathrm{Great}\:\mathrm{sirs} \\ $$
Answered by FongXD last updated on 26/Sep/22
Answered by FongXD last updated on 26/Sep/22
Answered by FongXD last updated on 26/Sep/22
Commented by mr W last updated on 26/Sep/22
$${thanks}\:{sir}! \\ $$
Answered by mr W last updated on 26/Sep/22
![let a_n =b_n +a+bn+c5^n [b_(n+2) −5b_(n+1) +6b_n ]+[a+b(n+2)−5a−5b(n+1)+6a+6bn]+[c5^(n+2) −5c5^(n+1) +6c5^n ]=3n+5^n [b_(n+2) −5b_(n+1) +6b_n ]+[2a−3b+2bn]+[6c5^n ]=3n+5^n 6c=1 ⇒c=(1/6) 2b=3 ⇒b=(3/2) 2a−3b=0 ⇒a=((3b)/2)=(9/4) b_(n+2) −5b_(n+1) +6b_n =0 r^2 −5r+6=0 r=2, 3 b_n =A2^n +B3^n ⇒a_n =A2^n +B3^n +(9/4)+((3n)/2)+(5^n /6) a_1 =A2+B3+(9/4)+(3/2)+(5/6)=1 ⇒2A+3B=−((43)/(12)) a_2 =A4+B9+(9/4)+(6/2)+((25)/6)=0 ⇒4A+9B=−((113)/(12)) ⇒A=−(2/3) ⇒B=−(3/4) ⇒a_n =−(2^(n+1) /3)−(3^(n+1) /4)+(9/4)+((3n)/2)+(5^n /6)](https://www.tinkutara.com/question/Q176770.png)
$${let}\:{a}_{{n}} ={b}_{{n}} +{a}+{bn}+{c}\mathrm{5}^{{n}} \\ $$$$\left[{b}_{{n}+\mathrm{2}} −\mathrm{5}{b}_{{n}+\mathrm{1}} +\mathrm{6}{b}_{{n}} \right]+\left[{a}+{b}\left({n}+\mathrm{2}\right)−\mathrm{5}{a}−\mathrm{5}{b}\left({n}+\mathrm{1}\right)+\mathrm{6}{a}+\mathrm{6}{bn}\right]+\left[{c}\mathrm{5}^{{n}+\mathrm{2}} −\mathrm{5}{c}\mathrm{5}^{{n}+\mathrm{1}} +\mathrm{6}{c}\mathrm{5}^{{n}} \right]=\mathrm{3}{n}+\mathrm{5}^{{n}} \\ $$$$\left[{b}_{{n}+\mathrm{2}} −\mathrm{5}{b}_{{n}+\mathrm{1}} +\mathrm{6}{b}_{{n}} \right]+\left[\mathrm{2}{a}−\mathrm{3}{b}+\mathrm{2}{bn}\right]+\left[\mathrm{6}{c}\mathrm{5}^{{n}} \right]=\mathrm{3}{n}+\mathrm{5}^{{n}} \\ $$$$\mathrm{6}{c}=\mathrm{1}\:\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\mathrm{2}{b}=\mathrm{3}\:\Rightarrow{b}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}{a}−\mathrm{3}{b}=\mathrm{0}\:\Rightarrow{a}=\frac{\mathrm{3}{b}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${b}_{{n}+\mathrm{2}} −\mathrm{5}{b}_{{n}+\mathrm{1}} +\mathrm{6}{b}_{{n}} =\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{5}{r}+\mathrm{6}=\mathrm{0} \\ $$$${r}=\mathrm{2},\:\mathrm{3} \\ $$$${b}_{{n}} ={A}\mathrm{2}^{{n}} +{B}\mathrm{3}^{{n}} \\ $$$$\Rightarrow{a}_{{n}} ={A}\mathrm{2}^{{n}} +{B}\mathrm{3}^{{n}} +\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{3}{n}}{\mathrm{2}}+\frac{\mathrm{5}^{{n}} }{\mathrm{6}} \\ $$$${a}_{\mathrm{1}} ={A}\mathrm{2}+{B}\mathrm{3}+\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{6}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{A}+\mathrm{3}{B}=−\frac{\mathrm{43}}{\mathrm{12}} \\ $$$${a}_{\mathrm{2}} ={A}\mathrm{4}+{B}\mathrm{9}+\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{6}}{\mathrm{2}}+\frac{\mathrm{25}}{\mathrm{6}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{A}+\mathrm{9}{B}=−\frac{\mathrm{113}}{\mathrm{12}} \\ $$$$\Rightarrow{A}=−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{B}=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{a}_{{n}} =−\frac{\mathrm{2}^{{n}+\mathrm{1}} }{\mathrm{3}}−\frac{\mathrm{3}^{{n}+\mathrm{1}} }{\mathrm{4}}+\frac{\mathrm{9}}{\mathrm{4}}+\frac{\mathrm{3}{n}}{\mathrm{2}}+\frac{\mathrm{5}^{{n}} }{\mathrm{6}} \\ $$
Commented by Tawa11 last updated on 26/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$