- #1
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($d_n)$
[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($d_n)$
The $d_n$ notation represents the domain of the function, or the input values that the function can take. In this case, the function is defined as $100 + n^2$, where n represents the input value.
To find the maximum value of the function, we can take the derivative of the function and set it equal to 0. This will give us the critical points of the function, and we can then determine which point corresponds to the maximum value. In this case, the maximum value will occur at n=0, with a value of 100.
The function $100 + n^2$ is a simple quadratic function that can be used to model a variety of real-world phenomena. It can be used to represent anything from the growth of a population to the trajectory of a projectile. In scientific research, this function can be used as a starting point for more complex models or to analyze data and make predictions.
As n increases, the maximum value of the function will also increase. This is because the function $n^2$ is an increasing function, meaning that as the input value increases, the output value also increases. Therefore, as n increases, the value of $100 + n^2$ will also increase, resulting in a higher maximum value.
No, the maximum value of the function $100 + n^2$ will never be negative. This is because the function has a minimum value of 100, which occurs at n=0. As n increases, the value of the function will only increase, never reaching a negative value. Therefore, the maximum value of this function will always be a positive number.