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 A sequence by recursion $a_{n+1} = a_n + \sqrt{a_{n+1}+a_n}$
07-27-2010, 03:25 PM
Post: #1
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
A sequence by recursion $a_{n+1} = a_n + \sqrt{a_{n+1}+a_n}$
Let $a_1 = 1, \; a_{n+1} = a_n + \sqrt{a_{n+1}+a_n}$
Find the explicit formula for $a_n$
07-28-2010, 02:36 PM (This post was last modified: 08-03-2010 02:15 AM by elim.)
Post: #2
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
RE: A sequence by recursion $a_{n+1} = a_n + \sqrt{a_{n+1}+a_n}$
$1 = a_1 = a_0 + \sqrt{a_0+a_1} = a_0 + \sqrt{1+a_0} \Rightarrow a_0 = 0$
$(a_{n+1}-a_n)-(a_n-a_{n-1}) = \sqrt{a_{n+1}+a_n}-\sqrt{a_n-a_{n-1}} = \frac{a_{n+1}-a_{n-1}}{\sqrt{a_{n+1}+a_n}+\sqrt{a_n-a_{n-1}} }=\frac{a_{n+1}-a_{n-1}}{(a_{n+1}-a_n)+(a_n-a_{n-1}) }=1$
Let $b_n = a_n - a_{n-1}$ then $b_1 = 1$ and $b_{n+1} = b_n +1$, therefore $b_n = n$ and so $a_n = \sum_{k=1}^n (a_k-a_{k-1}) = \sum_{k=1}^n k = n(n+1)/2$
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