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Question Number 176708 by Matica last updated on 25/Sep/22

a_(n+2) −3a_(n+1) +2a_n =n+3^n please find a_n

$$\:{a}_{{n}+\mathrm{2}} −\mathrm{3}{a}_{{n}+\mathrm{1}} +\mathrm{2}{a}_{{n}} ={n}+\mathrm{3}^{{n}} \: \\ $$$${please}\:{find}\:{a}_{{n}} \\ $$$$ \\ $$

Answered by mr W last updated on 26/Sep/22

(a_(n+2) −a_(n+1) )−2(a_(n+1) −a_n )=n+3^n let b_n =a_(n+1) −a_n b_(n+1) −2b_n =n+3^n let b_n =c_n +p+qn+3^n k b_(n+1) =c_(n+1) +p+q(n+1)+3^(n+1) k c_(n+1) −2c_n +p+q(n+1)−2p−2qn+3^(n+1) k−3^n k=n+3^n c_(n+1) −2c_n +q−p−qn+2k3^n =n+3^n ⇒q=−1 ⇒q−p=0 ⇒p=−1 ⇒2k=1 ⇒k=(1/2) ⇒b_n =c_n −1−n+(3^n /2) ⇒c_(n+1) −2c_n =0 ⇒c_n =2c_(n−1) =2^2 c_(n−2) =...=2^n c_0 c_0 =b_0 +1+0+(3^0 /2)=b_0 +(3/2)=a_1 −a_0 +(3/2) ⇒b_n =a_(n+1) −a_n =2^n c_0 −1−n+(3^n /2) Σ_(n=1) ^n (a_(n+1) −a_n )=Σ_(n=1) ^n (2^n c_0 −1−n+(3^n /2)) a_(n+1) −a_1 =((2c_0 (2^n −1))/(2−1))−n−((n(n+1))/2)+((3(3^n −1))/(2(3−1))) a_(n+1) =a_1 +2c_0 (2^n −1)−((n(n+3))/2)+((3^(n+1) −3)/4) ⇒a_n =a_1 +(a_1 −a_0 +(3/2))(2^n −2)−(((n−1)(n+2))/2)+((3^n −3)/4)

$$\:\left({a}_{{n}+\mathrm{2}} −{a}_{{n}+\mathrm{1}} \right)−\mathrm{2}\left({a}_{{n}+\mathrm{1}} −{a}_{{n}} \right)={n}+\mathrm{3}^{{n}} \\ $$$${let}\:{b}_{{n}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} \: \\ $$$${b}_{{n}+\mathrm{1}} −\mathrm{2}{b}_{{n}} ={n}+\mathrm{3}^{{n}} \: \\ $$$${let}\:{b}_{{n}} ={c}_{{n}} +{p}+{qn}+\mathrm{3}^{{n}} {k} \\ $$$${b}_{{n}+\mathrm{1}} ={c}_{{n}+\mathrm{1}} +{p}+{q}\left({n}+\mathrm{1}\right)+\mathrm{3}^{{n}+\mathrm{1}} {k} \\ $$$${c}_{{n}+\mathrm{1}} −\mathrm{2}{c}_{{n}} +{p}+{q}\left({n}+\mathrm{1}\right)−\mathrm{2}{p}−\mathrm{2}{qn}+\mathrm{3}^{{n}+\mathrm{1}} {k}−\mathrm{3}^{{n}} {k}={n}+\mathrm{3}^{{n}} \\ $$$${c}_{{n}+\mathrm{1}} −\mathrm{2}{c}_{{n}} +{q}−{p}−{qn}+\mathrm{2}{k}\mathrm{3}^{{n}} ={n}+\mathrm{3}^{{n}} \\ $$$$\Rightarrow{q}=−\mathrm{1} \\ $$$$\Rightarrow{q}−{p}=\mathrm{0}\:\Rightarrow{p}=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{k}=\mathrm{1}\:\Rightarrow{k}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{b}_{{n}} ={c}_{{n}} −\mathrm{1}−{n}+\frac{\mathrm{3}^{{n}} }{\mathrm{2}} \\ $$$$\Rightarrow{c}_{{n}+\mathrm{1}} −\mathrm{2}{c}_{{n}} =\mathrm{0} \\ $$$$\Rightarrow{c}_{{n}} =\mathrm{2}{c}_{{n}−\mathrm{1}} =\mathrm{2}^{\mathrm{2}} {c}_{{n}−\mathrm{2}} =...=\mathrm{2}^{{n}} {c}_{\mathrm{0}} \\ $$$${c}_{\mathrm{0}} ={b}_{\mathrm{0}} +\mathrm{1}+\mathrm{0}+\frac{\mathrm{3}^{\mathrm{0}} }{\mathrm{2}}={b}_{\mathrm{0}} +\frac{\mathrm{3}}{\mathrm{2}}={a}_{\mathrm{1}} −{a}_{\mathrm{0}} +\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{b}_{{n}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} =\mathrm{2}^{{n}} {c}_{\mathrm{0}} −\mathrm{1}−{n}+\frac{\mathrm{3}^{{n}} }{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left({a}_{{n}+\mathrm{1}} −{a}_{{n}} \right)=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}^{{n}} {c}_{\mathrm{0}} −\mathrm{1}−{n}+\frac{\mathrm{3}^{{n}} }{\mathrm{2}}\right) \\ $$$${a}_{{n}+\mathrm{1}} −{a}_{\mathrm{1}} =\frac{\mathrm{2}{c}_{\mathrm{0}} \left(\mathrm{2}^{{n}} −\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}−{n}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}+\frac{\mathrm{3}\left(\mathrm{3}^{{n}} −\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{3}−\mathrm{1}\right)} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{\mathrm{1}} +\mathrm{2}{c}_{\mathrm{0}} \left(\mathrm{2}^{{n}} −\mathrm{1}\right)−\frac{{n}\left({n}+\mathrm{3}\right)}{\mathrm{2}}+\frac{\mathrm{3}^{{n}+\mathrm{1}} −\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{a}_{{n}} ={a}_{\mathrm{1}} +\left({a}_{\mathrm{1}} −{a}_{\mathrm{0}} +\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\mathrm{2}^{{n}} −\mathrm{2}\right)−\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{3}^{{n}} −\mathrm{3}}{\mathrm{4}} \\ $$

Commented by Matica last updated on 26/Sep/22

Thank you very much

$${Thank}\:{you}\:{very}\:{much} \\ $$

Commented by Tawa11 last updated on 25/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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