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Question Number 76207 by mr W last updated on 25/Dec/19
if a_1 =1 and a_(n+1) =3a_n +n^2 find a_n =?
$${if}\:{a}_{\mathrm{1}} =\mathrm{1}\:{and}\:{a}_{{n}+\mathrm{1}} =\mathrm{3}{a}_{{n}} +{n}^{\mathrm{2}} \\ $$$${find}\:{a}_{{n}} =? \\ $$
Answered by mind is power last updated on 25/Dec/19
a_(n+1) =3a_n +n^2 a_n =f3^(n−1) +c_n c_(n+1) =3c_n +n^2 c_n =an^2 +bn+t a(n+1)^2 +b(n+1)+t=3an^2 +n^2 +3bn+3t a=3a+1 2a+b=3b a+b=2t a=−(1/2),b=−(1/2) t=−(1/2) a_n =f3^(n−1) −(1/2)(n^2 +n+1) a_1 =f−(3/2)=1⇒f=(5/2) a_n =(1/2)(5.3^(n−1) −n^2 −n−1)
$$\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\mathrm{3a}_{\mathrm{n}} +\mathrm{n}^{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{f3}^{\mathrm{n}−\mathrm{1}} +\mathrm{c}_{\mathrm{n}} \\ $$$$\mathrm{c}_{\mathrm{n}+\mathrm{1}} =\mathrm{3c}_{\mathrm{n}} +\mathrm{n}^{\mathrm{2}} \\ $$$$\mathrm{c}_{\mathrm{n}} =\mathrm{an}^{\mathrm{2}} +\mathrm{bn}+\mathrm{t} \\ $$$$\mathrm{a}\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{b}\left(\mathrm{n}+\mathrm{1}\right)+\mathrm{t}=\mathrm{3an}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} +\mathrm{3bn}+\mathrm{3t} \\ $$$$\mathrm{a}=\mathrm{3a}+\mathrm{1} \\ $$$$\mathrm{2a}+\mathrm{b}=\mathrm{3b} \\ $$$$\mathrm{a}+\mathrm{b}=\mathrm{2t} \\ $$$$\mathrm{a}=−\frac{\mathrm{1}}{\mathrm{2}},\mathrm{b}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{t}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{f3}^{\mathrm{n}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}\right) \\ $$$$\mathrm{a}_{\mathrm{1}} =\mathrm{f}−\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{1}\Rightarrow\mathrm{f}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}.\mathrm{3}^{\mathrm{n}−\mathrm{1}} −\mathrm{n}^{\mathrm{2}} −\mathrm{n}−\mathrm{1}\right) \\ $$$$ \\ $$
Answered by john santuy last updated on 25/Dec/19
a_2 = 3.1+1=4 a_3 =3×4+2^2 =16 a_4 =3×16+3^2 =57 etc....
$${a}_{\mathrm{2}} \:=\:\mathrm{3}.\mathrm{1}+\mathrm{1}=\mathrm{4} \\ $$$${a}_{\mathrm{3}} =\mathrm{3}×\mathrm{4}+\mathrm{2}^{\mathrm{2}} =\mathrm{16} \\ $$$${a}_{\mathrm{4}} =\mathrm{3}×\mathrm{16}+\mathrm{3}^{\mathrm{2}} =\mathrm{57} \\ $$$${etc}…. \\ $$
Commented by mr W last updated on 25/Dec/19
i got a_n =((5×3^(n−1) −n^2 −n−1)/2)
$${i}\:{got}\: \\ $$$${a}_{{n}} =\frac{\mathrm{5}×\mathrm{3}^{{n}−\mathrm{1}} −{n}^{\mathrm{2}} −{n}−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by john santuy last updated on 25/Dec/19
how sir?
$${how}\:{sir}? \\ $$
Answered by mr W last updated on 25/Dec/19
my attempt: a_(n+2) =3a_(n+1) +(n+1)^2 a_(n+1) =3a_n +n^2 ⇒a_(n+2) −a_(n+1) =3(a_(n+1) −a_n )+2n+1 let b_(n+2) =a_(n+2) −a_(n+1) ⇒b_(n+2) =3b_(n+1) +2n+1 assume b_n =c_n +pn+q with p, q as constants c_(n+2) +p(n+2)+q=3[c_(n+1) +p(n+1)+q]+2n+1 ⇒c_(n+2) =3c_(n+1) +2(p+1)n+(p+2q+1) now we select p and q such that the colored terms become zero: p+1=0 ⇒ p=−1 p+2q+1=0 ⇒−1+2q+1=0 ⇒q=0 i.e. b_n =c_n −n or c_n =b_n +n from c_(n+2) =3c_(n+1) we see c_n is a G.P. with common ratio 3. ⇒c_n =3^(n−2) c_2 a_1 =1 a_2 =3×1+1^2 =4 b_2 =a_2 −a_1 =4−1=3 c_2 =b_2 +2=3+2=5 ⇒c_n =5×3^(n−2) ⇒b_n =5×3^(n−2) −n ⇒b_(k+1) =5×3^(k−1) −k−1 ⇒a_(k+1) −a_k =5×3^(k−1) −k−1 ⇒Σ_(k=1) ^n a_(k+1) −Σ_(k=1) ^n a_k =5×Σ_(k=1) ^n 3^(k−1) −Σ_(k=1) ^n k−n ⇒a_(n+1) −a_1 =5×((3^n −1)/(3−1))−((n(n+1))/2)−n ⇒a_(n+1) =((5(3^n −1)−n(n+1)−2(n−1))/2) or ⇒a_n =((5(3^(n−1) −1)−(n−1)n−2(n−2))/2) ⇒a_n =((5×3^(n−1) −n^2 −n−1)/2)
$${my}\:{attempt}: \\ $$$${a}_{{n}+\mathrm{2}} =\mathrm{3}{a}_{{n}+\mathrm{1}} +\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${a}_{{n}+\mathrm{1}} =\mathrm{3}{a}_{{n}} +{n}^{\mathrm{2}} \\ $$$$\Rightarrow{a}_{{n}+\mathrm{2}} −{a}_{{n}+\mathrm{1}} =\mathrm{3}\left({a}_{{n}+\mathrm{1}} −{a}_{{n}} \right)+\mathrm{2}{n}+\mathrm{1} \\ $$$${let}\:\boldsymbol{{b}}_{\boldsymbol{{n}}+\mathrm{2}} =\boldsymbol{{a}}_{\boldsymbol{{n}}+\mathrm{2}} −\boldsymbol{{a}}_{\boldsymbol{{n}}+\mathrm{1}} \\ $$$$\Rightarrow{b}_{{n}+\mathrm{2}} =\mathrm{3}{b}_{{n}+\mathrm{1}} +\mathrm{2}{n}+\mathrm{1} \\ $$$${assume}\:\boldsymbol{{b}}_{\boldsymbol{{n}}} =\boldsymbol{{c}}_{\boldsymbol{{n}}} +\boldsymbol{{pn}}+\boldsymbol{{q}}\:{with}\:{p},\:{q}\:{as}\:{constants} \\ $$$${c}_{{n}+\mathrm{2}} +{p}\left({n}+\mathrm{2}\right)+{q}=\mathrm{3}\left[{c}_{{n}+\mathrm{1}} +{p}\left({n}+\mathrm{1}\right)+{q}\right]+\mathrm{2}{n}+\mathrm{1} \\ $$$$\Rightarrow{c}_{{n}+\mathrm{2}} =\mathrm{3}{c}_{{n}+\mathrm{1}} +\mathrm{2}\left({p}+\mathrm{1}\right){n}+\left({p}+\mathrm{2}{q}+\mathrm{1}\right) \\ $$$${now}\:{we}\:{select}\:{p}\:{and}\:{q}\:{such}\:{that}\:{the} \\ $$$${colored}\:{terms}\:{become}\:{zero}: \\ $$$${p}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:{p}=−\mathrm{1} \\ $$$${p}+\mathrm{2}{q}+\mathrm{1}=\mathrm{0}\:\Rightarrow−\mathrm{1}+\mathrm{2}{q}+\mathrm{1}=\mathrm{0}\:\Rightarrow{q}=\mathrm{0} \\ $$$${i}.{e}.\:\boldsymbol{{b}}_{\boldsymbol{{n}}} =\boldsymbol{{c}}_{\boldsymbol{{n}}} −\boldsymbol{{n}}\:{or}\:\boldsymbol{{c}}_{\boldsymbol{{n}}} =\boldsymbol{{b}}_{\boldsymbol{{n}}} +\boldsymbol{{n}} \\ $$$${from}\:{c}_{{n}+\mathrm{2}} =\mathrm{3}{c}_{{n}+\mathrm{1}} \:{we}\:{see}\:{c}_{{n}} \:{is}\:{a}\:{G}.{P}. \\ $$$${with}\:{common}\:{ratio}\:\mathrm{3}. \\ $$$$\Rightarrow{c}_{{n}} =\mathrm{3}^{{n}−\mathrm{2}} {c}_{\mathrm{2}} \\ $$$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{3}×\mathrm{1}+\mathrm{1}^{\mathrm{2}} =\mathrm{4} \\ $$$${b}_{\mathrm{2}} ={a}_{\mathrm{2}} −{a}_{\mathrm{1}} =\mathrm{4}−\mathrm{1}=\mathrm{3} \\ $$$${c}_{\mathrm{2}} ={b}_{\mathrm{2}} +\mathrm{2}=\mathrm{3}+\mathrm{2}=\mathrm{5} \\ $$$$\Rightarrow\boldsymbol{{c}}_{\boldsymbol{{n}}} =\mathrm{5}×\mathrm{3}^{\boldsymbol{{n}}−\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{{b}}_{\boldsymbol{{n}}} =\mathrm{5}×\mathrm{3}^{\boldsymbol{{n}}−\mathrm{2}} −\boldsymbol{{n}} \\ $$$$\Rightarrow\boldsymbol{{b}}_{{k}+\mathrm{1}} =\mathrm{5}×\mathrm{3}^{{k}−\mathrm{1}} −{k}−\mathrm{1} \\ $$$$\Rightarrow\boldsymbol{{a}}_{{k}+\mathrm{1}} −\boldsymbol{{a}}_{{k}} =\mathrm{5}×\mathrm{3}^{{k}−\mathrm{1}} −{k}−\mathrm{1} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\boldsymbol{{a}}_{{k}+\mathrm{1}} −\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\boldsymbol{{a}}_{{k}} =\mathrm{5}×\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{3}^{{k}−\mathrm{1}} −\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}−{n} \\ $$$$\Rightarrow\boldsymbol{{a}}_{{n}+\mathrm{1}} −\boldsymbol{{a}}_{\mathrm{1}} =\mathrm{5}×\frac{\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{3}−\mathrm{1}}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−{n} \\ $$$$\Rightarrow\boldsymbol{{a}}_{{n}+\mathrm{1}} =\frac{\mathrm{5}\left(\mathrm{3}^{{n}} −\mathrm{1}\right)−{n}\left({n}+\mathrm{1}\right)−\mathrm{2}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${or} \\ $$$$\Rightarrow\boldsymbol{{a}}_{{n}} =\frac{\mathrm{5}\left(\mathrm{3}^{{n}−\mathrm{1}} −\mathrm{1}\right)−\left({n}−\mathrm{1}\right){n}−\mathrm{2}\left({n}−\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{{a}}_{{n}} =\frac{\mathrm{5}×\mathrm{3}^{{n}−\mathrm{1}} −{n}^{\mathrm{2}} −{n}−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by john santuy last updated on 25/Dec/19
wawww great sir
$${wawww}\:\:\:{great}\:{sir} \\ $$
Answered by vishalbhardwaj last updated on 25/Dec/19
a_n = a_(n+1) −a_n = 3a_n +n^2 −3a_(n−1) −(n−1)^2 = 3[a_n −a_(n−1) ]+n^2 −n^2 +2n−1 (i) =3×common difference+2n−1 now by the given a_(n+1) =3a_n +n^2 a_2 =3a_1 +1 and a_2 =3(1)+1=4 ⇒ common difference = a_2 −a_1 =3 ⇒ by equation (i), a_n = 3(3)+2n−1 ⇒ a_n = 2n+8 this is the solution when series an A.P.
$${a}_{{n}} \:=\:{a}_{{n}+\mathrm{1}} −{a}_{{n}} \\ $$$$=\:\mathrm{3}{a}_{{n}} +{n}^{\mathrm{2}} −\mathrm{3}{a}_{{n}−\mathrm{1}} −\left({n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\:\mathrm{3}\left[{a}_{{n}} −{a}_{{n}−\mathrm{1}} \right]+{n}^{\mathrm{2}} −{n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{1}\:\:\left(\mathrm{i}\right) \\ $$$$=\mathrm{3}×{common}\:{difference}+\mathrm{2}{n}−\mathrm{1} \\ $$$$\mathrm{now}\:\mathrm{by}\:\mathrm{the}\:\mathrm{given}\:{a}_{{n}+\mathrm{1}} =\mathrm{3}{a}_{{n}} +{n}^{\mathrm{2}} \\ $$$${a}_{\mathrm{2}} =\mathrm{3}{a}_{\mathrm{1}} +\mathrm{1}\:\:\mathrm{and}\:{a}_{\mathrm{2}} =\mathrm{3}\left(\mathrm{1}\right)+\mathrm{1}=\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{common}\:\mathrm{difference}\:=\:{a}_{\mathrm{2}} −{a}_{\mathrm{1}} =\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{by}\:\mathrm{equation}\:\left(\mathrm{i}\right),\:{a}_{{n}} =\:\mathrm{3}\left(\mathrm{3}\right)+\mathrm{2}{n}−\mathrm{1} \\ $$$$\Rightarrow\:{a}_{{n}} =\:\mathrm{2}{n}+\mathrm{8} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{when}\:\mathrm{series}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}. \\ $$
Commented by MJS last updated on 25/Dec/19
but obviously it is not an A.P.
$$\mathrm{but}\:\mathrm{obviously}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}. \\ $$

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