Hello wizard654zzz Originally Posted by
wizard654zzz Let $\displaystyle (a_n) $be the sequence defined $\displaystyle a_n = [\frac{n^2 + 8n +10}{n+9}]$ where (x) is the largest integer that does not exceed x, Find the value of $\displaystyle \sum_{n=1}^{30}a_n$ ??
Divide and get quotient and remainder:
$\displaystyle \frac{n^2 + 8n +10}{n+9}=(n-1) + \frac{19}{n+9}$
Now for $\displaystyle 1\le n\le 10, \;2>\frac{19}{n+9}\ge 1$. Hence $\displaystyle a_n = n$.
But for $\displaystyle n > 10, \;\frac{19}{n+9}<1$. Hence $\displaystyle a_n = n-1$.
So the answer is:
$\displaystyle 1+2+3+...+10+10+11+12+...+29 = 445$
Grandad