Menu Close

if-a-n-5a-n-1-6a-n-2-1-and-a-0-1-a-1-2-find-a-n-in-terms-of-n-

Tinku Tara 0 Comments
Pin




Question Number 140102 by mr W last updated on 04/May/21
if a_n −5a_(n−1) +6a_(n−2) =1 and a_0 =1, a_1 =2. find a_n in terms of n.
$${if}\:{a}_{{n}} −\mathrm{5}{a}_{{n}−\mathrm{1}} +\mathrm{6}{a}_{{n}−\mathrm{2}} =\mathrm{1}\:{and}\: \\ $$$${a}_{\mathrm{0}} =\mathrm{1},\:{a}_{\mathrm{1}} =\mathrm{2}. \\ $$$${find}\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$
Commented by som(math1967) last updated on 04/May/21
Is it ((3^n +1)/2) sir ?
$${Is}\:{it}\:\frac{\mathrm{3}^{{n}} +\mathrm{1}}{\mathrm{2}}\:{sir}\:? \\ $$
Commented by mr W last updated on 04/May/21
yes, correct.
$${yes},\:{correct}. \\ $$
Commented by mr W last updated on 04/May/21
not correct sir! just check a_2 , a_3 . they should be: a_2 −5×2+6×1=1 ⇒a_2 =5 a_3 −5×5+6×2=1 ⇒a_3 =14
$${not}\:{correct}\:{sir}! \\ $$$${just}\:{check}\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} .\:{they}\:{should}\:{be}: \\ $$$${a}_{\mathrm{2}} −\mathrm{5}×\mathrm{2}+\mathrm{6}×\mathrm{1}=\mathrm{1}\:\Rightarrow{a}_{\mathrm{2}} =\mathrm{5} \\ $$$${a}_{\mathrm{3}} −\mathrm{5}×\mathrm{5}+\mathrm{6}×\mathrm{2}=\mathrm{1}\:\Rightarrow{a}_{\mathrm{3}} =\mathrm{14} \\ $$
Commented by som(math1967) last updated on 04/May/21
a_n −5a_(n−1) +6a_(n−2) =1 n=2 a_2 =1+5×2−6=5 a_3 =14....a_4 =41 a_0 ,a_1 ,a_2 ,a_3 1,(1+3^0 ),(1+3^0 +3^1 ),(1+3^0 +3+3^2 ) ....(1+3^0 +3+3^2 +....)_(a_n ) a_n =1+((3^n −1)/(3−1))=((3^n +1)/2)
$${a}_{{n}} −\mathrm{5}{a}_{{n}−\mathrm{1}} +\mathrm{6}{a}_{{n}−\mathrm{2}} =\mathrm{1} \\ $$$${n}=\mathrm{2} \\ $$$${a}_{\mathrm{2}} =\mathrm{1}+\mathrm{5}×\mathrm{2}−\mathrm{6}=\mathrm{5} \\ $$$${a}_{\mathrm{3}} =\mathrm{14}….{a}_{\mathrm{4}} =\mathrm{41} \\ $$$${a}_{\mathrm{0}} ,{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} \\ $$$$\mathrm{1},\left(\mathrm{1}+\mathrm{3}^{\mathrm{0}} \right),\left(\mathrm{1}+\mathrm{3}^{\mathrm{0}} +\mathrm{3}^{\mathrm{1}} \right),\left(\mathrm{1}+\mathrm{3}^{\mathrm{0}} +\mathrm{3}+\mathrm{3}^{\mathrm{2}} \right) \\ $$$$….\underset{{a}_{{n}} } {\underbrace{\left(\mathrm{1}+\mathrm{3}^{\mathrm{0}} +\mathrm{3}+\mathrm{3}^{\mathrm{2}} +….\right)}} \\ $$$${a}_{{n}} =\mathrm{1}+\frac{\mathrm{3}^{{n}} −\mathrm{1}}{\mathrm{3}−\mathrm{1}}=\frac{\mathrm{3}^{{n}} +\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 04/May/21
this method is not general enough. if i had given a_1 =1, a_2 =3, it would not be so easy to get the solution using this method. anyway, thanks alot!
$${this}\:{method}\:{is}\:{not}\:{general}\:{enough}. \\ $$$${if}\:{i}\:{had}\:{given}\:{a}_{\mathrm{1}} =\mathrm{1},\:{a}_{\mathrm{2}} =\mathrm{3},\:{it}\:{would} \\ $$$${not}\:{be}\:{so}\:{easy}\:{to}\:{get}\:{the}\:{solution}\:{using} \\ $$$${this}\:{method}. \\ $$$${anyway},\:{thanks}\:{alot}! \\ $$
Commented by Dwaipayan Shikari last updated on 04/May/21
Σa_n x^n −5Σa_(n−1) x^n +6Σa_(n−2) x^n =Σx^n ⇒Σ_(n=0) ^∞ a_(n+2) x^n −5Σ_(n=0) ^∞ a_(n+1) x^n +6Σ_(n=0) ^∞ a_n x^n =Σx^n ⇒(a_2 x+a_3 x^2 +..)−5(a_1 x+a_2 x^2 +...)+6(a_0 +a_1 x+a_2 x^2 +...)=(1/(1−x)) ⇒(1/x)(a_0 +a_1 x+a_2 x^2 +...)−(a_0 /x)−a_1 +(a_0 +a_1 x^2 +..)+5a_0 =(1/(1−x)) ⇒(1/x)Λ(x)−(1/x)−2+Λ(x)+5=(1/(1−x)) ⇒Λ(x)(((x+1)/x))=(1/x)+(1/(1−x))−3 ⇒Λ(x)=(1/(1+x))+(x/((1−x)(1+x)))−((3x)/((1+x))).. We have to turn it in a Taylor series
$$\Sigma{a}_{{n}} {x}^{{n}} −\mathrm{5}\Sigma{a}_{{n}−\mathrm{1}} {x}^{{n}} +\mathrm{6}\Sigma{a}_{{n}−\mathrm{2}} {x}^{{n}} =\Sigma{x}^{{n}} \\ $$$$\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}+\mathrm{2}} {x}^{{n}} −\mathrm{5}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}+\mathrm{1}} {x}^{{n}} +\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {x}^{{n}} =\Sigma{x}^{{n}} \\ $$$$\Rightarrow\left({a}_{\mathrm{2}} {x}+{a}_{\mathrm{3}} {x}^{\mathrm{2}} +..\right)−\mathrm{5}\left({a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…\right)+\mathrm{6}\left({a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}\left({a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +…\right)−\frac{{a}_{\mathrm{0}} }{{x}}−{a}_{\mathrm{1}} +\left({a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}^{\mathrm{2}} +..\right)+\mathrm{5}{a}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}\Lambda\left({x}\right)−\frac{\mathrm{1}}{{x}}−\mathrm{2}+\Lambda\left({x}\right)+\mathrm{5}=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow\Lambda\left({x}\right)\left(\frac{{x}+\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{1}−{x}}−\mathrm{3} \\ $$$$\Rightarrow\Lambda\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{{x}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}−\frac{\mathrm{3}{x}}{\left(\mathrm{1}+{x}\right)}.. \\ $$$${We}\:{have}\:{to}\:{turn}\:{it}\:{in}\:{a}\:{Taylor}\:{series} \\ $$
Commented by mr W last updated on 04/May/21
thanks to all for trying different ways!
$${thanks}\:{to}\:{all}\:{for}\:{trying}\:{different}\:{ways}! \\ $$
Answered by mr W last updated on 04/May/21
such that it′s not too special, i′ll change the condition to a_0 =1, a_1 =3. a_n −5a_(n−1) +6a_(n−2) =1 at first we shall try to eliminate the term 1 on the RHS. let a_n =b_n +k, k is a constant. then we will get b_n +k−5(b_(n−1) +k)+6(b_(n−2) +k)=1 b_n −5b_(n−1) +6b_(n−2) =1−2k we set 1−2k=0, i.e. k=(1/2), then b_n −5b_(n−1) +6b_(n−2) =0 let b_n =Ap^n Ap^n −5Ap^(n−1) +6Ap^(n−2) =0 Ap^(n−2) (p^2 −5p+6)=0 since A≠0, p≠0, ⇒p^2 −5p+6=0 ⇒(p−2)(p−3)=0 ⇒p=2 or 3 b_n can be generally expressed as b_n =A2^n +B3^n with A,B =constants ⇒a_n =A2^n +B3^n +(1/2) a_0 =A+B+(1/2)=1 (given) ...(i) a_1 =A×2+B×3+(1/2)=3 (given) ...(ii) from (i) and (ii) we get ⇒A=−1 ⇒B=(3/2) ⇒a_n =−2^n +(3/2)×3^n +(1/2) ⇒a_n =((3^(n+1) +1)/2)−2^n ■
$${such}\:{that}\:{it}'{s}\:{not}\:{too}\:{special},\:{i}'{ll} \\ $$$${change}\:{the}\:{condition}\:{to} \\ $$$${a}_{\mathrm{0}} =\mathrm{1},\:{a}_{\mathrm{1}} =\mathrm{3}. \\ $$$$ \\ $$$${a}_{{n}} −\mathrm{5}{a}_{{n}−\mathrm{1}} +\mathrm{6}{a}_{{n}−\mathrm{2}} =\mathrm{1} \\ $$$${at}\:{first}\:{we}\:{shall}\:{try}\:{to}\:{eliminate}\:{the} \\ $$$${term}\:\mathrm{1}\:{on}\:{the}\:{RHS}.\: \\ $$$${let}\:{a}_{{n}} ={b}_{{n}} +{k},\:{k}\:{is}\:{a}\:{constant}. \\ $$$${then}\:{we}\:{will}\:{get} \\ $$$${b}_{{n}} +{k}−\mathrm{5}\left({b}_{{n}−\mathrm{1}} +{k}\right)+\mathrm{6}\left({b}_{{n}−\mathrm{2}} +{k}\right)=\mathrm{1} \\ $$$${b}_{{n}} −\mathrm{5}{b}_{{n}−\mathrm{1}} +\mathrm{6}{b}_{{n}−\mathrm{2}} =\mathrm{1}−\mathrm{2}{k} \\ $$$${we}\:{set}\:\mathrm{1}−\mathrm{2}{k}=\mathrm{0},\:{i}.{e}.\:{k}=\frac{\mathrm{1}}{\mathrm{2}},\:{then} \\ $$$${b}_{{n}} −\mathrm{5}{b}_{{n}−\mathrm{1}} +\mathrm{6}{b}_{{n}−\mathrm{2}} =\mathrm{0} \\ $$$${let}\:{b}_{{n}} ={Ap}^{{n}} \\ $$$${Ap}^{{n}} −\mathrm{5}{Ap}^{{n}−\mathrm{1}} +\mathrm{6}{Ap}^{{n}−\mathrm{2}} =\mathrm{0} \\ $$$${Ap}^{{n}−\mathrm{2}} \left({p}^{\mathrm{2}} −\mathrm{5}{p}+\mathrm{6}\right)=\mathrm{0} \\ $$$${since}\:{A}\neq\mathrm{0},\:{p}\neq\mathrm{0}, \\ $$$$\Rightarrow{p}^{\mathrm{2}} −\mathrm{5}{p}+\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\left({p}−\mathrm{2}\right)\left({p}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{p}=\mathrm{2}\:{or}\:\mathrm{3} \\ $$$${b}_{{n}} \:{can}\:{be}\:{generally}\:{expressed}\:{as} \\ $$$${b}_{{n}} ={A}\mathrm{2}^{{n}} +{B}\mathrm{3}^{{n}} \\ $$$${with}\:{A},{B}\:={constants} \\ $$$$\Rightarrow{a}_{{n}} ={A}\mathrm{2}^{{n}} +{B}\mathrm{3}^{{n}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}_{\mathrm{0}} ={A}+{B}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1}\:\left({given}\right)\:\:\:\:…\left({i}\right) \\ $$$${a}_{\mathrm{1}} ={A}×\mathrm{2}+{B}×\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{3}\:\left({given}\right)\:\:\:…\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get} \\ $$$$\Rightarrow{A}=−\mathrm{1} \\ $$$$\Rightarrow{B}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{a}_{{n}} =−\mathrm{2}^{{n}} +\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{3}^{{n}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{a}_{{n}} =\frac{\mathrm{3}^{{n}+\mathrm{1}} +\mathrm{1}}{\mathrm{2}}−\mathrm{2}^{{n}} \\ $$$$\blacksquare \\ $$
Commented by BHOOPENDRA last updated on 04/May/21
thanks alot sir
$${thanks}\:{alot}\:{sir} \\ $$
Answered by floor(10²Eta[1]) last updated on 05/May/21
a_n =(((5−(√5))/(10)))(((5+(√5))/2))^n +(((5+(√5))/(10)))(((5−(√5))/2))^n
$$\mathrm{a}_{\mathrm{n}} =\left(\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{10}}\right)\left(\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{n}} +\left(\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{10}}\right)\left(\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{n}} \\ $$
Commented by mr W last updated on 05/May/21
wrong. a_2 =15, but should be 14.
$${wrong}. \\ $$$${a}_{\mathrm{2}} =\mathrm{15},\:{but}\:{should}\:{be}\:\mathrm{14}. \\ $$
Commented by floor(10²Eta[1]) last updated on 05/May/21
wrong. a_2 −5a_1 +6a_0 =1 a_2 =5
$$\mathrm{wrong}. \\ $$$$\mathrm{a}_{\mathrm{2}} −\mathrm{5a}_{\mathrm{1}} +\mathrm{6a}_{\mathrm{0}} =\mathrm{1} \\ $$$$\mathrm{a}_{\mathrm{2}} =\mathrm{5} \\ $$
Commented by mr W last updated on 05/May/21
a typo. i meant a_3 should be 14, not 15.
$${a}\:{typo}.\:{i}\:{meant}\:{a}_{\mathrm{3}} \:{should}\:{be}\:\mathrm{14},\:{not}\:\mathrm{15}. \\ $$
Commented by floor(10²Eta[1]) last updated on 05/May/21
you′re right
$$\mathrm{you}'\mathrm{re}\:\mathrm{right}\: \\ $$
Answered by floor(10²Eta[1]) last updated on 05/May/21
a_n =x_n +y_n a_n =5a_(n−1) −6a_(n−2) +1 x_n =5x_n −6x_n +1 x_n =(1/2) y_n =5y_(n−1) −6y_(n−2) y^2 −5y+6=0 y_1 =2, y_2 =3 y_n =c_1 2^n +c_2 3^n y_0 =c_1 +c_2 =a_0 −(1/2)=(1/2) y_1 =2c_1 +3c_2 =a_1 −(1/2)=(3/2) c_2 =(1/2)−c_1 2c_1 +(3/2)−3c_1 =(3/2)⇒c_1 =0⇒c_2 =(1/2) y_n =(3^n /2) ⇒a_n =((3^n +1)/2)
$$\mathrm{a}_{\mathrm{n}} =\mathrm{x}_{\mathrm{n}} +\mathrm{y}_{\mathrm{n}} \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{5a}_{\mathrm{n}−\mathrm{1}} −\mathrm{6a}_{\mathrm{n}−\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{x}_{\mathrm{n}} =\mathrm{5x}_{\mathrm{n}} −\mathrm{6x}_{\mathrm{n}} +\mathrm{1} \\ $$$$\mathrm{x}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{y}_{\mathrm{n}} =\mathrm{5y}_{\mathrm{n}−\mathrm{1}} −\mathrm{6y}_{\mathrm{n}−\mathrm{2}} \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{5y}+\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{y}_{\mathrm{1}} =\mathrm{2},\:\mathrm{y}_{\mathrm{2}} =\mathrm{3} \\ $$$$\mathrm{y}_{\mathrm{n}} =\mathrm{c}_{\mathrm{1}} \mathrm{2}^{\mathrm{n}} +\mathrm{c}_{\mathrm{2}} \mathrm{3}^{\mathrm{n}} \\ $$$$\mathrm{y}_{\mathrm{0}} =\mathrm{c}_{\mathrm{1}} +\mathrm{c}_{\mathrm{2}} =\mathrm{a}_{\mathrm{0}} −\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{y}_{\mathrm{1}} =\mathrm{2c}_{\mathrm{1}} +\mathrm{3c}_{\mathrm{2}} =\mathrm{a}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{c}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{c}_{\mathrm{1}} \\ $$$$\mathrm{2c}_{\mathrm{1}} +\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{3c}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow\mathrm{c}_{\mathrm{1}} =\mathrm{0}\Rightarrow\mathrm{c}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{y}_{\mathrm{n}} =\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{3}^{\mathrm{n}} +\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 05/May/21
thanks sir!
$${thanks}\:{sir}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *