1. ## help on sequence

I cannot get the correct answer of this problem. The answer is 445 but I got 435.

May be I didn't understand what the last sentence meant.

2. Originally Posted by shrrikesh
I cannot get the correct answer of this problem. The answer is 445 but I got 435.

May be I didn't understand what the last sentence meant.
It is a trick.
Find these values: $a_{10},~a_{11}~\&~a_{12}$.
Do you now see where the extra 10 comes from?

3. Originally Posted by shrrikesh
I cannot get the correct answer of this problem. The answer is 445 but I got 435.

May be I didn't understand what the last sentence meant.
I got 435 too - perhaps they're somehow rounding it differently?

edit: found it... those dastardly so and sos...

4. Originally Posted by Unenlightened
I got 435 too - perhaps they're somehow rounding it differently?
edit: plato - did you get a different answer?
$a_{10}=10,~a_{11}=10,~a_{12}=11$
That is 30 integers 1 to 29 with two 10's.

5. I feel cheated

6. 445 is correct.
If n=10: |190/19| = 10
If n=11: |219/20| = 10

Missed one of those?

Your answer of 435 is sum of digits from 1 to 29 ; 29*(30/2) = 435.
So you're missing a term, since you need 30 terms.

7. Originally Posted by shrrikesh
I cannot get the correct answer of this problem. The answer is 445 but I got 435.

May be I didn't understand what the last sentence meant.
First, make a simple conversion

$\sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{{n^2} + 8n + 10}}{{n + 9}}} \right\rfloor } = \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{n\left( {n + 9} \right) + 10 - n}}{{n + 9}}} \right\rfloor } = \sum\limits_{n = 1}^{30} {\left\lfloor {n + \frac{{10 - n}}{{n + 9}}} \right\rfloor } =$

$= \sum\limits_{n = 1}^{30} n + \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } = 465 + \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } .$

$\frac{{10 - n}}{{n + 9}} = 0 \Leftrightarrow n = 10$

Therefore, we consider two cases:

1) if $n = \left\{ {1,{\text{ }}2,{\text{ }} \ldots ,{\text{ }}10} \right\}$ then $\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor = 0$.

2) if $n = \left\{ {11,{\text{ 1}}2,{\text{ }} \ldots ,{\text{ }}30} \right\}$ then $\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor = - 1$.

So we have

$\sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } = \sum\limits_{n = 1}^{10} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } + \sum\limits_{n = 11}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } = 0 + \left( { - 20} \right) = - 20$.

Finally we have

$\sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{{n^2} + 8n + 10}}{{n + 9}}} \right\rfloor } = \sum\limits_{n = 1}^{30} n + \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } = 465 + \left( { - 20} \right) = 445.$