nth term of the sequence help

• Oct 30th 2012, 08:28 PM
gfbrd
nth term of the sequence help
Hey I need some help with this math problem.

Find a formula for the nth term of the sequence
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, ...

I can see the pattern, but I'm not sure how to go about finding a formula for this sequence. Hoping someone can guide me through this problem so I can understand it better thanks.
• Oct 30th 2012, 10:07 PM
Salahuddin559
Re: nth term of the sequence help
The term in the interval (n(n-1)/2, n(n+1)/2]is n. The pattern generalizes as follows: The first term is 1, the next two terms is 2, etc... (as you already know). So, every number (k) repeats k times, and the successive terms increase by 1, after repetition, starting from 1. It means, the nth term has to do with how many numbers are below it, and since each number repeats itself "its value" number of times, this has to do with sum of first n numbers. Once you cross, sum of first n numbers terms, you get a new value for the term in the series.

Salahuddin
Maths online
• Oct 31st 2012, 12:21 AM
gfbrd
Re: nth term of the sequence help
sorry, i already understand how the pattern works, what im looking for is someone to guide me through the problem to find a formula for this sequence, thanks for your help anyways
• Oct 31st 2012, 12:32 AM
Salahuddin559
Re: nth term of the sequence help
The term in the interval (n(n-1)/2, n(n+1)/2]is n. Doesnt that give a formula?

Salahuddin
Maths online
• Oct 31st 2012, 12:40 PM
Plato
Re: nth term of the sequence help
Quote:

Originally Posted by gfbrd
Hey I need some help with this math problem.
Find a formula for the nth term of the sequence
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, ...
I can see the pattern, but I'm not sure how to go about finding a formula for this sequence. Hoping someone can guide me through this problem so I can understand it better thanks.

This involves triangular numbers. The rule uses the floor function.
$\displaystyle s_n = \left\lfloor {\frac{{1 + \sqrt {8n - 7} }}{2}} \right\rfloor$.
• Oct 31st 2012, 08:28 PM
gfbrd
Re: nth term of the sequence help

Edit: Plato do you think you can explain to me how you get that answer?
• Nov 1st 2012, 05:38 AM
Plato
Re: nth term of the sequence help
Quote:

Originally Posted by gfbrd
The triangular numbers $\displaystyle \frac{n(n+1)}{2}+1,~~2,~4,~7,~11,~16,~\cdots$ are places in your sequence where the terms "change". That is, from the $\displaystyle 10^{th}+1$ term to the $\displaystyle 15^{th}$ term the sequence terms are $\displaystyle 5$ and $\displaystyle n=4$ gives the fourth triangular number. The floor function gives an integer. From terms $\displaystyle 11\text{ to }15$ $\displaystyle \left\lfloor {\frac{{1 + \sqrt {8n - 7} }}{2}} \right\rfloor=5$ at term $\displaystyle 16$ it changes to $\displaystyle 6$ for the next six terms.