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.

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

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

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

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 $.

Re: nth term of the sequence help

Thanks guys for your help

Edit: Plato do you think you can explain to me how you get that answer?

Re: nth term of the sequence help

Quote:

Originally Posted by

**gfbrd** Thanks guys for your help

Edit: Plato do you think you can explain to me how you get that answer?

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.

Re: nth term of the sequence help

oh alright thanks a lot plato