Question Number 8347 by sou1618 last updated on 09/Oct/16
$${a}_{\mathrm{1}} =\mathrm{2}\:,\:\:{a}_{{n}+\mathrm{1}} >{a}_{{n}} \\ $$$$\left({a}_{{n}+\mathrm{1}} −{a}_{{n}} \right)^{\mathrm{2}} =\:\mathrm{2}\left({a}_{{n}+\mathrm{1}} +{a}_{{n}} \right) \\ $$$$\:{a}_{{n}} =?? \\ $$$${help}\:{me}\:{please}. \\ $$
Commented by sou1618 last updated on 09/Oct/16
$${a}_{{n}+\mathrm{1}} ={a}_{{n}} +\mathrm{1}+\sqrt{\mathrm{4}{a}_{{n}} +\mathrm{1}} \\ $$$${a}_{\mathrm{1}} =\mathrm{2} \\ $$$${a}_{\mathrm{2}} =\mathrm{2}+\mathrm{1}+\sqrt{\mathrm{8}+\mathrm{1}}=\mathrm{6} \\ $$$${a}_{\mathrm{3}} =\mathrm{6}+\mathrm{1}+\sqrt{\mathrm{24}+\mathrm{1}}=\mathrm{12} \\ $$$${a}_{\mathrm{4}} =\mathrm{12}+\mathrm{1}+\sqrt{\mathrm{48}+\mathrm{1}}=\mathrm{20} \\ $$$${a}_{\mathrm{5}} =\mathrm{20}+\mathrm{1}+\sqrt{\mathrm{80}+\mathrm{1}}=\mathrm{30} \\ $$$$… \\ $$$${a}_{\mathrm{2}} −{a}_{\mathrm{1}} =\mathrm{4} \\ $$$${a}_{\mathrm{3}} −{a}_{\mathrm{2}} =\mathrm{6} \\ $$$${a}_{\mathrm{4}} −{a}_{\mathrm{3}} =\mathrm{8} \\ $$$${a}_{\mathrm{5}} −{a}_{\mathrm{4}} =\mathrm{10} \\ $$$$… \\ $$$$ \\ $$$${a}_{{n}+\mathrm{1}} −{a}_{{n}} =\mathrm{2}\left({n}+\mathrm{1}\right) \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} +\mathrm{2}{n}+\mathrm{2}\:\:,\:\:{a}_{\mathrm{1}} =\mathrm{2} \\ $$$${so} \\ $$$${a}_{{n}} =\mathrm{2}+\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\mathrm{2}{n}+\mathrm{2}\right) \\ $$$${a}_{{n}} =\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}}{n}\left({n}−\mathrm{1}\right)+\mathrm{2}\left({n}−\mathrm{1}\right) \\ $$$${a}_{{n}} ={n}\left({n}−\mathrm{1}\right)+\mathrm{2}{n} \\ $$$${a}_{{n}} ={n}\left({n}+\mathrm{1}\right) \\ $$$$ \\ $$$${isn}'{t}\:{it}\:{strict}? \\ $$
Commented by Rasheed Soomro last updated on 12/Oct/16
$$\left({a}_{{n}+\mathrm{1}} −{a}_{{n}} \right)^{\mathrm{2}} =\:\mathrm{2}\left({a}_{{n}+\mathrm{1}} +{a}_{{n}} \right)\overset{?} {\Rightarrow}{a}_{{n}+\mathrm{1}} ={a}_{{n}} +\mathrm{1}+\sqrt{\mathrm{4}{a}_{{n}} +\mathrm{1}} \\ $$$$\mathrm{Please}\:\mathrm{insert}\:\mathrm{some}\:\mathrm{steps}. \\ $$$$ \\ $$
Commented by prakash jain last updated on 14/Oct/16
![(a_(n+1) −a_n )^2 −2(a_(n+1) −a_n )+1−4a_n −1=0 (a_(n+1) −a_n −1)^2 =4a_n +1 a_(n+1) =1+a_n +(√(4a_n +1)) If a_n is a polynomial then 4a_n +1 is a perfect square of polynomial. Let 4a_n +1=(c_0 +c_1 n+c_2 n^2 +c_3 n^3 +...+c_k n^k )^2 a_n =[(c_0 +c_1 n+c_2 n^2 +c_3 n^3 +...+c_k n^k )^2 −1]/4 a_(n+1) =[(c_0 +c_1 (n+1)+c_2 (n+1)^2 +c_3 (n+1)^3 +...)^2 −1]/4 Try for 4a_n +1=(c_0 +c_1 n+c_2 n^2 )^2 a_n =(1/4)[(c_0 +c_1 n+c_2 n^2 )^2 −1] Equating coeffcients n^4 : (1/4)c_2 ^2 =(1/4)c_2 ^2 n^3 :c_2 ^2 +(1/2)c_1 c_2 =(1/2)c_1 c_2 ⇒c_2 =0 No polynomial solution is possible if 4a_n +1 is a square of degree higher than 1. Degree 1 4a_n +1=(c_0 +c_1 n)^2 n^2 :(1/4)c_1 ^2 =(1/4)c_1 ^2 n: (1/2)c_1 c_0 +(1/2)c_1 ^2 =(1/2)c_1 c_0 +c_1 ⇒c_1 =2 n^0 :(c_0 ^2 /4)+((c_1 c_0 )/2)+(c_1 ^2 /4)−(1/4)=(c_0 ^2 /4)+c_0 +(3/4) ⇒c_0 +1−1/4=c_0 +3/4 a_n =((c_0 +2n)^2 −1)/4 a_1 =2⇒(c_0 +2)^2 −1=8 c_0 ^2 +4c_0 +4−1=8 c_0 ^2 +4c_0 −5=0 c_0 ^2 +5c_0 −c_0 −5=0 (c_0 −1)(c_0 +5)^2 =0 a_n = { ((((1+2n)^2 −1)/4=n^2 +n)),((((−5+2n)^2 −1)/4=n^2 −5n+6=(n−2)(n−3))) :} check (a_(n+1) −a_n )^2 =[(n−1)(n−2)−(n−2)(n−3)]^2 =4(n−2)^2 2(a_(n+1) +a_n )=2(n−2)(2n−4)=4(n−2)^2 comparing results determinant ((n,(n(n+1)),((n−2)(n−3))),(1,2,2),(2,6,0),(3,(12),0),(4,(20),2),(5,(30),6))](https://www.tinkutara.com/question/Q8509.png)
$$\left({a}_{{n}+\mathrm{1}} −{a}_{{n}} \right)^{\mathrm{2}} −\mathrm{2}\left({a}_{{n}+\mathrm{1}} −{a}_{{n}} \right)+\mathrm{1}−\mathrm{4}{a}_{{n}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({a}_{{n}+\mathrm{1}} −{a}_{{n}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}{a}_{{n}} +\mathrm{1} \\ $$$${a}_{{n}+\mathrm{1}} =\mathrm{1}+{a}_{{n}} +\sqrt{\mathrm{4}{a}_{{n}} +\mathrm{1}} \\ $$$$\mathrm{If}\:{a}_{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{then}\:\mathrm{4}{a}_{{n}} +\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect} \\ $$$$\mathrm{square}\:\mathrm{of}\:\mathrm{polynomial}. \\ $$$$\mathrm{Let}\:\mathrm{4}{a}_{{n}} +\mathrm{1}=\left({c}_{\mathrm{0}} +{c}_{\mathrm{1}} {n}+{c}_{\mathrm{2}} {n}^{\mathrm{2}} +{c}_{\mathrm{3}} {n}^{\mathrm{3}} +…+{c}_{{k}} {n}^{{k}} \right)^{\mathrm{2}} \\ $$$${a}_{{n}} =\left[\left({c}_{\mathrm{0}} +{c}_{\mathrm{1}} {n}+{c}_{\mathrm{2}} {n}^{\mathrm{2}} +{c}_{\mathrm{3}} {n}^{\mathrm{3}} +…+{c}_{{k}} {n}^{{k}} \right)^{\mathrm{2}} −\mathrm{1}\right]/\mathrm{4} \\ $$$${a}_{{n}+\mathrm{1}} =\left[\left({c}_{\mathrm{0}} +{c}_{\mathrm{1}} \left({n}+\mathrm{1}\right)+{c}_{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} +{c}_{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} +…\right)^{\mathrm{2}} −\mathrm{1}\right]/\mathrm{4} \\ $$$$\mathrm{Try}\:\mathrm{for}\: \\ $$$$\mathrm{4}{a}_{{n}} +\mathrm{1}=\left({c}_{\mathrm{0}} +{c}_{\mathrm{1}} {n}+{c}_{\mathrm{2}} {n}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${a}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}\left[\left({c}_{\mathrm{0}} +{c}_{\mathrm{1}} {n}+{c}_{\mathrm{2}} {n}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{1}\right] \\ $$$$\mathrm{Equating}\:\mathrm{coeffcients}\: \\ $$$${n}^{\mathrm{4}} :\:\frac{\mathrm{1}}{\mathrm{4}}{c}_{\mathrm{2}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}{c}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${n}^{\mathrm{3}} :{c}_{\mathrm{2}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{c}_{\mathrm{1}} {c}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{c}_{\mathrm{1}} {c}_{\mathrm{2}} \Rightarrow{c}_{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{No}\:\mathrm{polynomial}\:\mathrm{solution}\:\mathrm{is} \\ $$$$\mathrm{possible}\:\mathrm{if}\:\mathrm{4}{a}_{{n}} +\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{higher} \\ $$$$\mathrm{than}\:\mathrm{1}. \\ $$$$\mathrm{Degree}\:\mathrm{1} \\ $$$$\mathrm{4a}_{\mathrm{n}} +\mathrm{1}=\left(\mathrm{c}_{\mathrm{0}} +\mathrm{c}_{\mathrm{1}} \mathrm{n}\right)^{\mathrm{2}} \\ $$$$\mathrm{n}^{\mathrm{2}} :\frac{\mathrm{1}}{\mathrm{4}}\mathrm{c}_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\mathrm{c}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\mathrm{n}:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{c}_{\mathrm{1}} \mathrm{c}_{\mathrm{0}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{c}_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{c}_{\mathrm{1}} \mathrm{c}_{\mathrm{0}} +\mathrm{c}_{\mathrm{1}} \Rightarrow\mathrm{c}_{\mathrm{1}} =\mathrm{2} \\ $$$$\mathrm{n}^{\mathrm{0}} :\frac{\mathrm{c}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{c}_{\mathrm{1}} \mathrm{c}_{\mathrm{0}} }{\mathrm{2}}+\frac{\mathrm{c}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{c}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}}+\mathrm{c}_{\mathrm{0}} +\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{c}_{\mathrm{0}} +\mathrm{1}−\mathrm{1}/\mathrm{4}=\mathrm{c}_{\mathrm{0}} +\mathrm{3}/\mathrm{4} \\ $$$$\mathrm{a}_{\mathrm{n}} =\left(\left(\mathrm{c}_{\mathrm{0}} +\mathrm{2n}\right)^{\mathrm{2}} −\mathrm{1}\right)/\mathrm{4} \\ $$$$\mathrm{a}_{\mathrm{1}} =\mathrm{2}\Rightarrow\left(\mathrm{c}_{\mathrm{0}} +\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{8} \\ $$$$\mathrm{c}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{4c}_{\mathrm{0}} +\mathrm{4}−\mathrm{1}=\mathrm{8} \\ $$$$\mathrm{c}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{4c}_{\mathrm{0}} −\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{c}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{5c}_{\mathrm{0}} −\mathrm{c}_{\mathrm{0}} −\mathrm{5}=\mathrm{0} \\ $$$$\left(\mathrm{c}_{\mathrm{0}} −\mathrm{1}\right)\left(\mathrm{c}_{\mathrm{0}} +\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{a}_{\mathrm{n}} =\begin{cases}{\left(\left(\mathrm{1}+\mathrm{2n}\right)^{\mathrm{2}} −\mathrm{1}\right)/\mathrm{4}=\mathrm{n}^{\mathrm{2}} +\mathrm{n}}\\{\left(\left(−\mathrm{5}+\mathrm{2n}\right)^{\mathrm{2}} −\mathrm{1}\right)/\mathrm{4}=\mathrm{n}^{\mathrm{2}} −\mathrm{5n}+\mathrm{6}=\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right)}\end{cases} \\ $$$$\mathrm{check} \\ $$$$\left(\mathrm{a}_{\mathrm{n}+\mathrm{1}} −\mathrm{a}_{\mathrm{n}} \right)^{\mathrm{2}} =\left[\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{2}\right)−\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right)\right]^{\mathrm{2}} \\ $$$$=\mathrm{4}\left(\mathrm{n}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{a}_{\mathrm{n}+\mathrm{1}} +\mathrm{a}_{\mathrm{n}} \right)=\mathrm{2}\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{2n}−\mathrm{4}\right)=\mathrm{4}\left(\mathrm{n}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{comparing}\:\mathrm{results} \\ $$$$\begin{vmatrix}{\mathrm{n}}&{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}&{\left(\mathrm{n}−\mathrm{2}\right)\left(\mathrm{n}−\mathrm{3}\right)}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{2}}\\{\mathrm{2}}&{\mathrm{6}}&{\mathrm{0}}\\{\mathrm{3}}&{\mathrm{12}}&{\mathrm{0}}\\{\mathrm{4}}&{\mathrm{20}}&{\mathrm{2}}\\{\mathrm{5}}&{\mathrm{30}}&{\mathrm{6}}\end{vmatrix} \\ $$
Commented by prakash jain last updated on 14/Oct/16
$$\mathrm{The}\:\mathrm{second}\:\mathrm{solution}\:\mathrm{does}\:\mathrm{not}\:\mathrm{meet}\:\mathrm{the} \\ $$$$\mathrm{condition}\:\mathrm{that}\:{a}_{{n}+\mathrm{1}} >{a}_{{n}} \\ $$
Commented by sou1618 last updated on 15/Oct/16
$${a}_{{n}+\mathrm{1}} ^{\:\mathrm{2}} −\mathrm{2}{a}_{{n}+\mathrm{1}} {a}_{{n}} +{a}_{{n}} ^{\:\mathrm{2}} −\mathrm{2}{a}_{{n}+\mathrm{1}} −\mathrm{2}{a}_{{n}} =\mathrm{0} \\ $$$${a}_{{n}+\mathrm{1}} ^{\:\mathrm{2}} −\mathrm{2}\left({a}_{{n}} +\mathrm{1}\right){a}_{{n}+\mathrm{1}} +{a}_{{n}} ^{\:\mathrm{2}} −\mathrm{2}{a}_{{n}} =\mathrm{0} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} +\mathrm{1}\pm\sqrt{\left({a}_{{n}} +\mathrm{1}\right)^{\mathrm{2}} −\left({a}_{{n}} ^{\:\mathrm{2}} −\mathrm{2}{a}_{{n}} \right)} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} +\mathrm{1}\pm\sqrt{\mathrm{4}{a}_{{n}} +\mathrm{1}} \\ $$$$\ast\:{a}_{{n}+\mathrm{1}} >{a}_{{n}} \\ $$$${so} \\ $$$${a}_{{n}+\mathrm{1}} ={a}_{{n}} +\mathrm{1}\sqrt{\mathrm{4}{a}_{{n}} +\mathrm{1}} \\ $$$${thanks}. \\ $$$${I}\:{have}\:{to}\:{study}\:{hard}…. \\ $$