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$a_n$ is the sequence defined by $$a_n = \left[\frac{n^2+8n+10}{n+9}\right]$$ where $[x]$ is the largest integer that doesn't exceed $x$. Find $$\sum_{n=1}^{30} a_n$$

" Where $[x]$ is the largest integer that doesn't exceed $x$. " what does it mean? I haven't tried much, since i don't even understand the question...

$a_1= 19/10;\; a_2 = 30/13;\;a_3 = 43/12;,\ldots$

Khayyam
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Dini
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  • $\frac{n^2 + 8n +10}{n+9}= n-1 + \frac{19}{n+9}$ . So the thing behave as n for n less than 11 and n-1 for n greater than 10 – aryan bansal Jan 31 '20 at 17:11
  • $a_1 \ne \frac {19}{10}$. $\frac {19}{10}= 1.9$ and $1.9$ is not an integer. $...-1< 0< 1< \frac {19}{10} < 2< 3<...$ so $a_1 = [\frac {19}{10} ]=1$ which is the largest integer that does not exceed $1.9$. The integers $2$ and larger all exceed $1.9$ and the integers $1$ and less do not. $1$ is the largest integer that does not exceed $1.9$ so $[1.9] = 1$. – fleablood Jan 31 '20 at 17:13

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The function $[\;]$ which called "floor function" or "integer part" is defined as follows: For a real $x$, write $x$ in the form $x=n+f$ where $n$ is an integer and $f$ is so that $0\leq f<1$, then $[x]=n$. Specially, if $0\leq x<1$ then $[x]=0$. Also for negative numbers like $-3.14=-4+0.86$, thus $[-3.14]=-4$

For an answer to the question see If $a_n=\left[\frac{n^2+8n+10}{n+9}\right]$, find $ \sum_{n=1}^{30} a_n$

Qurultay
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  • In positive numbers [3.14] = 3. But [ -3.14] = -4? – Dini Jan 31 '20 at 17:16
  • since $-4 \le -3.14 < -3$ then all the integers $-3$ and larger exceed $-3.14$ and the largest integer that doesn't exceed is $-4$. So $[-3.14] = -4$. It means exactly what it says it means. – fleablood Jan 31 '20 at 17:18
  • For every real $x$, whether positive of negative, there is a unique integer $n$ so that $n \le x < n+1$ and $[x] = n$. If $x$ is negative then $n$ is negative. $|x| \ge |n|$. So yes, $[-3.14]$ is so that $-4 \le -3.14 < -3$ and $[-3.14] = -4$ and $|-3.14| < |-4|$. – fleablood Jan 31 '20 at 17:21