# Arithmetic Series

• May 1st 2008, 04:42 PM
stones44
Arithmetic Series
So there is wall...the top has 1 brick, then the next row has 3 bricks, then 5 bricks etc. there are 2500 bricks in the wall then how many rows are there

im just really confused with arithmetic series right now and need general help with them too....

pref. without using some formula but a more explanative version

• May 1st 2008, 04:45 PM
Mathstud29
Quote:

Originally Posted by stones44
So there is wall...the top has 1 brick, then the next row has 3 bricks, then 5 bricks etc. there are 2500 bricks in the wall then how many rows are there

im just really confused with arithmetic series right now and need general help with them too....

pref. without using some formula but a more explanative version

This would be a series of the odd numbers so it would be $\sum_{i=0}^{n}2i+1$

can you find n from that?
• May 1st 2008, 04:51 PM
stones44
no thats what im saying..im very confused
• May 1st 2008, 05:05 PM
Mathstud28
Quote:

Originally Posted by Mathstud29
This would be a series of the odd numbers so it would be $\sum_{i=0}^{n}2i+1$

can you find n from that?

$\sum_{n=0}^{x}2n+1=\sum_{n=0}^{x}2n+\sum_{n=0}^{x} 1=\frac{2x(x+1)}{2}+x$

Now solve $\frac{2x(x+1)}{2}+x=2500$
• May 1st 2008, 11:02 PM
flyingsquirrel
Hi
Quote:

Originally Posted by Mathstud28
$\sum_{n=0}^{x}2n+1=\sum_{n=0}^{x}2n+\sum_{n=0}^{x} 1=\frac{2x(x+1)}{2}+x{\color{red}+1}$

There are $n-p+1$ elements in the set of positive integers $\{p,\,p+1,\,p+2,\,\ldots,\,n-1,\,n\}$, assuming $n\geq p$.
• May 2nd 2008, 04:41 AM
Soroban
Hello, stones44!

Quote:

So there is wall.
The top has 1 brick, then the next row has 3 bricks, then 5 bricks, etc.
There are 2500 bricks in the wall.
How many rows are there?

Without a formula? . . . Then a child can do it!

Just start adding: . $1 + 3 + 5 + 7 + 9 + \hdots$ . until you reach 2500.

Then count up the number of numbers you added.

. . . I'll wait in the car . . .

Formula for the sum of the first $n$ terms of an arithmetic series:
. . $S_n \;=\;\frac{n }{2}[2a + (n-1)d]$

We have: .first term, $a = 1$ . . . common difference, $d = 2$

We know: . $S_n = 2500$ . . . and we want $n$.

So we have: . $2500 \:=\:\frac{n}{2}[2(1) + (n-1)2] \quad\Rightarrow\quad 2500 \:=\:\frac{n}{2}(2n)\quad\Rightarrow\quad n^2 \:=\:2500$

Therefore: . $n \:=\:50$ . . . There are 50 rows.