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 , then
So the problem is equivalent to finding k such as :
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
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
so it's going like this, so to me looks to be bigger number
anyway thanks very much for your help and your time
Merci
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
=>
Assuming that is in the form with P a polynomial of degree 2, you can try to take a recurrence formula from it, as you know
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...