# Geometric Series [Problem]

• Apr 8th 2008, 10:27 AM
bybuti
Geometric Series [Problem]
Hi everyone, this is my first post here :)

I would be glad if someone help me with this problem

1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,7....

a1000000= ?

a its geometric serie.

• Apr 8th 2008, 10:55 AM
Moo
Hello,

You can notice that :

if $a_n=k$, then $\sum_{i=1}^k i

So the problem is equivalent to finding k such as :

$\underbrace{\sum_{i=1}^k i}_{\dfrac{k(k+1)}{2}} < 1000000 < \underbrace{\sum_{i=1}^{k+1} i}_{\dfrac{(k+1)(k+2)}{2}}$

You can find k by the calculus, but i can give you an approximation :

if k=1413, k(k+1)/2=998991 and (k+1)(k+2)/2=1000405

So k=1413 is solution.

Hence $a_{1000000}=1413$

Err...actually, i don't know why you are told a is geometric series though the series they gave you is kinda arithmetical oO
Perhaps i made a mistake, so if it seems wrong to you, don't hesitate to tell me :s
• Apr 8th 2008, 11:56 AM
bybuti
hmm, it's looking small number that solve that you post

hmm, i think its like this 1+2+2+3+3+3+4+4+4+4+5+5+5+5+5+6+6+6+6+6+6+7+7+7+...+

as you can see this goes like this 1 is one time, number 2 its 2 times, number 3 three times , ... number 10 its repeated ten times and continues like this.

so from this up we have $a_{10}=30$

$a_{20}=85$

$a_{30}=156$

so it's going like this, so $a_{1000000}=$ to me looks to be bigger number :)

Merci :)
• Apr 8th 2008, 12:02 PM
Moo
Ha !

I thought that the series of $a_n$ was beginning with 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,7....

I'm sorry, but i can't tell you for the sum of all these terms (need to think more seriously about it)

(you're welcome :D)
• Apr 8th 2008, 12:04 PM
Moo
Oh, actually, it's simple...

1+2+2+3+3+3+4+4+4+4+5+5+5+5+5+6+6+6+6+6+6+7+7+7+.. .+

This is $\sum_{k=1}^n k*k = \sum_{k=1}^n k^2$

Well, after two lines, i manage to remove reason from my message... I don't remember the value for this sum...
• Apr 8th 2008, 12:04 PM
bybuti
hmm, now looks more logical :p

more near to solution :)
• Apr 8th 2008, 12:22 PM
Moo
I have a way to find it... Perhaps you can think a little about it (it's very hard to write in latex when ideas just come out...)

(k+1)²=k²+2k+1

=> $\sum_{k=1}^n (k+1)^2 = \sum_{k=1}^n k^2 +2\sum_{k=1}^n k + \sum_{k=1}^n 1$

Assuming that $\sum_{k=1}^n k^2$ is in the form $P(n)$ with P a polynomial of degree 2, you can try to take a recurrence formula from it, as you know $2\sum_{k=1}^n k + \sum_{k=1}^n 1$

Ok it seems that P is of degree 3.

Maxima gave me this :

(%i12) sum(k^2, k, 1, n), simpsum;
(%o12) (2*n^3+3*n^2+n)/6

& Good luck, because it's possible to find k when you have this formula, but it's very long to do it without calculator.
I'm pretty sure there is a more direct way for this problem, but i'm sorry i can't find any...
• Apr 8th 2008, 12:40 PM
Plato
Quote:

Originally Posted by bybuti
so it's going like this, so $a_{1000000}=$ to me looks to be bigger number.

It is a larger number. But, only by one: 1414.
Using the ceiling function, $a_n = \left\lceil {\frac{{ - 1 + \sqrt {1 + 8n} }}{2}} \right\rceil$.
• Apr 8th 2008, 12:43 PM
Moo
Gargl..

I hesitated between 1413 and 1414, i didn't remember why i took 1413 although my first result was 1414 :'( Can you help me please ? Now i'm wondering ~
• Apr 8th 2008, 01:04 PM
Plato
Quote:

Originally Posted by Moo
$a_n=k$, then $\sum_{i=1}^k i

Your mistake is right there. Using that scheme we get $a_n=4$ implies that $10 \le n \le 15$ but $a_{10} = 4,\,a_{11} = 5,\,a_{12} = 5,\,a_{13} = 5,\,a_{14} = 5,\,a_{15} = 5,\,a_{16} = 5$.
• Apr 8th 2008, 01:09 PM
Moo
Hm so the solution is k+1, not k ^^
Thanks
• Apr 8th 2008, 01:17 PM
Plato
Quote:

Originally Posted by Moo
Hm so the solution is k+1, not k

Just change it to
$a_n=k>1$, then $\sum_{i=1}^{k-1} i < n \le \sum_{i=1}^{k} i$
• Apr 8th 2008, 01:18 PM
Moo
Yeah this sounds better :-)
Now i have to find back why i changed my mind ~
• Apr 9th 2008, 08:50 AM
bybuti
Thanks guys, i will see if this is the wanted result :)