Albert1
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$ a_n=100+n^2.\,\, n=1,2,3,----$
$d_n=(a_n,a_{n+1})$
find :max($d_n)$
$d_n=(a_n,a_{n+1})$
find :max($d_n)$
The maximum value of \(d_n\) occurs at \(n = 200\) and is equal to 401. The sequence \(a_n\) is defined as \(a_n = 100 + n^2\) for \(n = 1, 2, 3, \ldots\). The relationship \(d_n = (a_n, a_{n+1})\) leads to the conclusion that \(d\) divides \(401\), which is confirmed by the calculations involving \(a_{200} = 40100\) and \(a_{201} = 40501\). Thus, the maximum \(d_n\) is definitively established as 401.
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[sp]If $d$ divides $a_n$ and $a_{n+1}$ then $d$ divides $a_{n+1} - a_n = 2n+1$. So $d$ divides $2a_n - n(2n+1) = 200-n$. Therefore $d$ divides $(2n+1) + 2(200-n) = 401$. But $401$ divides $a_{200}$ ($= 40100$) and $a_{201}$ ($= 40501$). Thus $\max d_n = 401.$[/sp]Albert said:$ a_n=100+n^2.\,\, n=1,2,3,----$
$d_n=(a_n,a_{n+1})$
find :max($d_n)$