Math Help - sequences and summations...i don't even know where to start...

1. sequences and summations...i don't even know where to start...

The homeworks are getting overwhelming...I have NO IDEA where to start...I haven't taken a math class in 10 years (calculus II).

Let a-sub-n be the nth term f the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6,..., constructed by including the integer k exactly k times. Show that a-sub-n = floor(sqrt(2n) + 1/2).

I honestly don't even know what the first step is or what I should even be thinking about...

2. This is like...the very first problem of only our 5th homework assignment...they're already too hard for me. Anyone?

3. Hello teamtrinity
Originally Posted by teamtrinity
The homeworks are getting overwhelming...I have NO IDEA where to start...I haven't taken a math class in 10 years (calculus II).

Let a-sub-n be the nth term f the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6,..., constructed by including the integer k exactly k times. Show that a-sub-n = floor(sqrt(2n) + 1/2).

I honestly don't even know what the first step is or what I should even be thinking about...
OK, here's where to start.

The last of the terms that has value $k-1$ is in $i^{th}$ position where:
$i = 1 + 2 + 3 + ... + (k-1) = \tfrac12k(k-1)$
and, similarly, the last of the integers with value $k$ is in the position $\tfrac12k(k+1)$.

So for $\tfrac12k(k-1) + 1\le n \le \tfrac12k(k+1),\; a_n = k$.

Can you continue?

how did you find out that that was the place to start? Can you explain a bit more, what's going on here? I see what you're saying. You spotted the relationship that:

1 = 1
2 = 3
3 = 6
4 = 10

but like teamtrinity, I would have started with the following
$\lfloor \sqrt{2n} + \frac{1}{2} \rfloor \Leftrightarrow x \leq \sqrt{2n} + \frac{1}{2} < x + 1$

and then probably not know how to go further! How did you work it out?

5. Hello bmp05
Originally Posted by bmp05
how did you find out that that was the place to start? Can you explain a bit more, what's going on here? I see what you're saying. You spotted the relationship that:

1 = 1
2 = 3
3 = 6
4 = 10

but like teamtrinity, I would have started with the following
$\lfloor \sqrt{2n} + \frac{1}{2} \rfloor \Leftrightarrow x \leq \sqrt{2n} + \frac{1}{2} < x + 1$

and then probably not know how to go further! How did you work it out?
First, let's just complete the proof. We've shown that the terms with value $k$ are in position $n$, where:
$\tfrac12k(k-1)+1\le n \le \tfrac12k(k+1)$

$\Rightarrow k(k-1) + 2 \le 2n \le k(k+1)$
Completing the square for the LH inequality:
$(k-\tfrac12)^2 +\tfrac74\le 2n$

$\Rightarrow k \le \sqrt{2n-\tfrac74}+\tfrac12$

$\Rightarrow k < \sqrt{2n} + \tfrac12$
And in the same way for the RH inequality, we get:
$2n\le (k+\tfrac12)^2-\tfrac14$
$\Rightarrow \sqrt{2n+\tfrac14} - \tfrac12 < k$

$\Rightarrow k +1> \sqrt{2n} +\tfrac12$
So, putting the two together:
$\Rightarrow k < \sqrt{2n} + \tfrac12

$\Rightarrow k =a_n= \lfloor\sqrt{2n} + \tfrac12\rfloor$
So how did I know that was the place to start? The honest answer is that I didn't. It was simply that this was the only way I could think of to describe what was happening. Clearly the position of the last one of any given set of integers, value $k$, was the sum of the first $k$ natural numbers which is well known to be $\tfrac12k(k+1)$. So it was just a question of working out where these integers would start, and writing all this down algebraically.

Until I just worked out the above proof of the actual result involving the floor function, I had no idea what would happen next. But if you begin by writing down an algebraic condition to describe something that you know to be true, then the result should follow automatically. But sometimes, of course, it still isn't that easy!