a_1 =2 , a_(n+1) >a_n (a_(n+1) −a_n )^2 = 2(a_(n+1) +a_n ) a


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Question Number 8347 by sou1618 last updated on 09/Oct/16

$a_{1} = 2, a_{n + 1} > a_{n}$ ${(a_{n + 1} - a_{n})}^{2} = 2 (a_{n + 1} + a_{n})$ $a_{n} = ? ?$ $h e l p m e p l e a s e .$
Commented by sou1618 last updated on 09/Oct/16

$a_{n + 1} = a_{n} + 1 + \sqrt{4 a_{n} + 1}$ $a_{1} = 2$ $a_{2} = 2 + 1 + \sqrt{8 + 1} = 6$ $a_{3} = 6 + 1 + \sqrt{24 + 1} = 12$ $a_{4} = 12 + 1 + \sqrt{48 + 1} = 20$ $a_{5} = 20 + 1 + \sqrt{80 + 1} = 30$ $. . .$ $a_{2} - a_{1} = 4$ $a_{3} - a_{2} = 6$ $a_{4} - a_{3} = 8$ $a_{5} - a_{4} = 10$ $. . .$ $a_{n + 1} - a_{n} = 2 (n + 1)$ $a_{n + 1} = a_{n} + 2 n + 2, a_{1} = 2$ $s o$ $a_{n} = 2 + \underset{k = 1}{\sum^{n - 1}} (2 n + 2)$ $a_{n} = 2 + \frac{2}{2} n (n - 1) + 2 (n - 1)$ $a_{n} = n (n - 1) + 2 n$ $a_{n} = n (n + 1)$ ${i s n}^{'} t i t s t r i c t ?$
Commented by Rasheed Soomro last updated on 12/Oct/16

${(a_{n + 1} - a_{n})}^{2} = 2 (a_{n + 1} + a_{n}) \overset{?}{\Rightarrow} a_{n + 1} = a_{n} + 1 + \sqrt{4 a_{n} + 1}$ $Please insert some steps .$
Commented by prakash jain last updated on 14/Oct/16
$(a_(n+1) −a_n )^2 −2(a_(n+1) −a_n )+1−4a_n −1=0 (a_(n+1) −a_n −1)^2 =4a_n +1 a_(n+1) =1+a_n +(√(4a_n +1)) If a_n is a polynomial then 4a_n +1 is a perfect square of polynomial. Let 4a_n +1=(c_0 +c_1 n+c_2 n^2 +c_3 n^3 +...+c_k n^k )^2 a_n =[(c_0 +c_1 n+c_2 n^2 +c_3 n^3 +...+c_k n^k )^2 −1]/4 a_(n+1) =[(c_0 +c_1 (n+1)+c_2 (n+1)^2 +c_3 (n+1)^3 +...)^2 −1]/4 Try for 4a_n +1=(c_0 +c_1 n+c_2 n^2 )^2 a_n =(1/4)[(c_0 +c_1 n+c_2 n^2 )^2 −1] Equating coeffcients n^4 : (1/4)c_2 ^2 =(1/4)c_2 ^2 n^3 :c_2 ^2 +(1/2)c_1 c_2 =(1/2)c_1 c_2 ⇒c_2 =0 No polynomial solution is possible if 4a_n +1 is a square of degree higher than 1. Degree 1 4a_n +1=(c_0 +c_1 n)^2 n^2 :(1/4)c_1 ^2 =(1/4)c_1 ^2 n: (1/2)c_1 c_0 +(1/2)c_1 ^2 =(1/2)c_1 c_0 +c_1 ⇒c_1 =2 n^0 :(c_0 ^2 /4)+((c_1 c_0 )/2)+(c_1 ^2 /4)−(1/4)=(c_0 ^2 /4)+c_0 +(3/4) ⇒c_0 +1−1/4=c_0 +3/4 a_n =((c_0 +2n)^2 −1)/4 a_1 =2⇒(c_0 +2)^2 −1=8 c_0 ^2 +4c_0 +4−1=8 c_0 ^2 +4c_0 −5=0 c_0 ^2 +5c_0 −c_0 −5=0 (c_0 −1)(c_0 +5)^2 =0 a_n = { ((((1+2n)^2 −1)/4=n^2 +n)),((((−5+2n)^2 −1)/4=n^2 −5n+6=(n−2)(n−3))) :} check (a_(n+1) −a_n )^2 =[(n−1)(n−2)−(n−2)(n−3)]^2 =4(n−2)^2 2(a_(n+1) +a_n )=2(n−2)(2n−4)=4(n−2)^2 comparing results determinant ((n,(n(n+1)),((n−2)(n−3))),(1,2,2),(2,6,0),(3,(12),0),(4,(20),2),(5,(30),6))$
${(a_{n + 1} - a_{n})}^{2} - 2 (a_{n + 1} - a_{n}) + 1 - 4 a_{n} - 1 = 0$ ${(a_{n + 1} - a_{n} - 1)}^{2} = 4 a_{n} + 1$ $a_{n + 1} = 1 + a_{n} + \sqrt{4 a_{n} + 1}$ $If a_{n} is a polynomial then 4 a_{n} + 1 is a perfect$ $square of polynomial .$ $Let 4 a_{n} + 1 = {(c_{0} + c_{1} n + c_{2} n^{2} + c_{3} n^{3} + . . . + c_{k} n^{k})}^{2}$ $a_{n} = [{(c_{0} + c_{1} n + c_{2} n^{2} + c_{3} n^{3} + . . . + c_{k} n^{k})}^{2} - 1] / 4$ $a_{n + 1} = [{(c_{0} + c_{1} (n + 1) + c_{2} {(n + 1)}^{2} + c_{3} {(n + 1)}^{3} + . . .)}^{2} - 1] / 4$ $Try for$ $4 a_{n} + 1 = {(c_{0} + c_{1} n + c_{2} n^{2})}^{2}$ $a_{n} = \frac{1}{4} [{(c_{0} + c_{1} n + c_{2} n^{2})}^{2} - 1]$ $Equating coeffcients$ $n^{4} : \frac{1}{4} c_{2}^{2} = \frac{1}{4} c_{2}^{2}$ $n^{3} : c_{2}^{2} + \frac{1}{2} c_{1} c_{2} = \frac{1}{2} c_{1} c_{2} \Rightarrow c_{2} = 0$ $No polynomial solution is$ $possible if 4 a_{n} + 1 is a square of degree higher$ $than 1 .$ $Degree 1$ ${4 a}_{n} + 1 = {(c_{0} + c_{1} n)}^{2}$ $n^{2} : \frac{1}{4} c_{1}^{2} = \frac{1}{4} c_{1}^{2}$ $n : \frac{1}{2} c_{1} c_{0} + \frac{1}{2} c_{1}^{2} = \frac{1}{2} c_{1} c_{0} + c_{1} \Rightarrow c_{1} = 2$ $n^{0} : \frac{c_{0}^{2}}{4} + \frac{c_{1} c_{0}}{2} + \frac{c_{1}^{2}}{4} - \frac{1}{4} = \frac{c_{0}^{2}}{4} + c_{0} + \frac{3}{4}$ $\Rightarrow c_{0} + 1 - 1 / 4 = c_{0} + 3 / 4$ $a_{n} = ({(c_{0} + 2 n)}^{2} - 1) / 4$ $a_{1} = 2 \Rightarrow {(c_{0} + 2)}^{2} - 1 = 8$ $c_{0}^{2} + {4 c}_{0} + 4 - 1 = 8$ $c_{0}^{2} + {4 c}_{0} - 5 = 0$ $c_{0}^{2} + {5 c}_{0} - c_{0} - 5 = 0$ $(c_{0} - 1) {(c_{0} + 5)}^{2} = 0$ $a_{n} = {\begin{cases} ({(1 + 2 n)}^{2} - 1) / 4 = n^{2} + n \\ ({(- 5 + 2 n)}^{2} - 1) / 4 = n^{2} - 5 n + 6 = (n - 2) (n - 3) \end{cases}$ $check$ ${(a_{n + 1} - a_{n})}^{2} = {[(n - 1) (n - 2) - (n - 2) (n - 3)]}^{2}$ $= 4 {(n - 2)}^{2}$ $2 (a_{n + 1} + a_{n}) = 2 (n - 2) (2 n - 4) = 4 {(n - 2)}^{2}$ $comparing results$ $\| \begin{matrix} n & n (n + 1) & (n - 2) (n - 3) \\ 1 & 2 & 2 \\ 2 & 6 & 0 \\ 3 & 12 & 0 \\ 4 & 20 & 2 \\ 5 & 30 & 6 \end{matrix} \|$
Commented by prakash jain last updated on 14/Oct/16

$The second solution does not meet the$ $condition that a_{n + 1} > a_{n}$
Commented by sou1618 last updated on 15/Oct/16

$a_{n + 1}^{2} - 2 a_{n + 1} a_{n} + a_{n}^{2} - 2 a_{n + 1} - 2 a_{n} = 0$ $a_{n + 1}^{2} - 2 (a_{n} + 1) a_{n + 1} + a_{n}^{2} - 2 a_{n} = 0$ $a_{n + 1} = a_{n} + 1 \pm \sqrt{{(a_{n} + 1)}^{2} - (a_{n}^{2} - 2 a_{n})}$ $a_{n + 1} = a_{n} + 1 \pm \sqrt{4 a_{n} + 1}$ $* a_{n + 1} > a_{n}$ $s o$ $a_{n + 1} = a_{n} + 1 \sqrt{4 a_{n} + 1}$ $t h a n k s .$ $I h a v e t o s t u d y h a r d . . . .$
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