a-1-2-a-n-1-gt-a-n-a-n-1-a-n-2-2-a-n-1-a-n-a-n-help-me-please-

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

\dots

a_{2} - a_{1} = 4

a_{3} - a_{2} = 6

a_{4} - a_{3} = 8

a_{5} - a_{4} = 10

\dots

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 - 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} + \dots + c_{k} n^{k})}^{2}

a_{n} = [{(c_{0} + c_{1} n + c_{2} n^{2} + c_{3} n^{3} + \dots + 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} + \dots)}^{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 \dots .

Facebook Tweet Pin