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.
thanks in advance
Hello,
You can notice that :
if $\displaystyle a_n=k$, then $\displaystyle \sum_{i=1}^k i<n<\sum_{i=1}^{k+1} i$
So the problem is equivalent to finding k such as :
$\displaystyle \underbrace{\sum_{i=1}^k i}_{\dfrac{k(k+1)}{2}} < 1000000 < \underbrace{\sum_{i=1}^{k+1} i}_{\dfrac{(k+1)(k+2)}{2}}$
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 $\displaystyle a_{1000000}=1413$
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
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 $\displaystyle a_{10}=30$
$\displaystyle a_{20}=85$
$\displaystyle a_{30}=156$
so it's going like this, so $\displaystyle a_{1000000}=$ to me looks to be bigger number
anyway thanks very much for your help and your time
Merci
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 $\displaystyle \sum_{k=1}^n k*k = \sum_{k=1}^n k^2$
Well, after two lines, i manage to remove reason from my message... I don't remember the value for this sum...
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
=> $\displaystyle \sum_{k=1}^n (k+1)^2 = \sum_{k=1}^n k^2 +2\sum_{k=1}^n k + \sum_{k=1}^n 1$
Assuming that $\displaystyle \sum_{k=1}^n k^2$ is in the form $\displaystyle P(n)$ with P a polynomial of degree 2, you can try to take a recurrence formula from it, as you know $\displaystyle 2\sum_{k=1}^n k + \sum_{k=1}^n 1$
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...