Arithmetic Series

stones44 · #1 $Old$ May 1st, 2008, 05:42 PM

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

Mathstud29 · #2 $Old$ May 1st, 2008, 05:45 PM

Quote:

Originally Posted by stones44 $View Post$

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?

stones44 · #3 $Old$ May 1st, 2008, 05:51 PM

no thats what im saying..im very confused

Mathstud28 · #4 $Old$ May 1st, 2008, 06:05 PM

Quote:

Originally Posted by Mathstud29 $View Post$

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 $\frac{2x(x+1)}{2}+x=2500$

flyingsquirrel · #5 $Old$ May 2nd, 2008, 12:02 AM

Hi

Quote:

Originally Posted by Mathstud28 $View Post$

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

There are $n-p+1$ elements in the set of positive integers $\{p,\,p+1,\,p+2,\,\ldots,\,n-1,\,n\}$ , assuming $n\geq p$ .

Soroban · #6 $Old$ May 2nd, 2008, 05:41 AM

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.