1. ## Geometric Progression

Hi... can anyone help me with this ?
looks simple, i know! but i forgot the method =(

basically i want to expand

the sum of r between r = 1 upto n-1

to get

1/2n(n-1)

how would i do this?

thanks!

2. Originally Posted by matlabnoob
Hi... can anyone help me with this ?
looks simple, i know! but i forgot the method =(

basically i want to expand

the sum of r between r = 1 upto n-1

to get

1/2n(n-1)

how would i do this?

thanks!
Hi matlabnoob,

notice that $1+2+3=(2-1)+2+(2+1)=2+2+2=3(2)$

and $1+2+3+4+5=(3-2)+(3-1)+3+(3+1)+(3+2)=3+3+3+3+3=5(3)$

Therefore, this is the same as $(number\ of\ terms)(average\ value)$

For a large number of terms, simply take the average to be $\frac{1st\ term+last\ term}{2}$

Therefore $\sum_{r=1}^{n-1}r=\frac{1+n-1}{2}(n-1)=\frac{n(n-1)}{2}$

3. thank you for your help!

i understood that. now im trying to do the same for ... r^2...

and im getting the same answer as for r

am i right to use the same method for r^2??

thanks again!=]

4. Doing what with r^2? You titled this "geometric progression" but the first example you gave was an arithmetic progression. What is it you are trying to do?

5. Originally Posted by matlabnoob

i understood that. now im trying to do the same for ... r^2...

and im getting the same answer as for r

am i right to use the same method for r^2??

thanks again!=]
No, you cannot use the same method.
The previous one shows how to develop a formula for summing consecutive natural numbers.

If you want to sum consecutive squares, that's more involved.

You would need to understand the previous one quite well before attempting
to find a quick way to sum squares.

Here is a way to do it using "telescoping".

If you want to sum squares from r=1 to n-1, then you could use the following

$(2r+1)^3-(2r-1)^3=(2r+1)(2r+1)(2r+1)-(2r-1)(2r-1)(2r-1)$

$=(4r^2+4r+1)(2r+1)-(4r^2-4r+1)(2r-1)=$ $(8r^3+4r^2+8r^2+4r+2r+1)-(8r^3-4r^2-8r^2+4r+2r-1)$

$=(8r^3+12r^2+6r+1)-(8r^3-12r^2+6r-1)=24r^2+2$

$\Rightarrow\ (2r+1)^3-(2r-1)^3=24r^2+2$

Telescoping

$r=1.......\ (2r+1)^3-(2r-1)^3=3^3-1^3$

$r=2.......\ (2r+1)^3-(2r-1)^3=5^3-3^3$

$r=3.......\ (2r+1)^3-(2r-1)^3=7^3-5^3$

$r=4.......\ (2r+1)^3-(2r-1)^3=9^3-7^3$

all the way to

$r=n-3....\ (2r+1)^3-(2r-1)^3=(2n-5)^3-(2n-7)^3$

$r=n-2....\ (2r+1)^3-(2r-1)^3=(2n-3)^3-(2n-5)^3$

$r=n-1....\ (2r+1)^3-(2r-1)^3=(2n-1)^3-(2n-3)^3$

When these are summed, all terms except $(2n-1)^3$ and 1 cancel.

Hence $\sum_{r=1}^{n-1}\left(24r^2+2\right)=(2n-1)^3-1$

2 summed n-1 times is 2(n-1)

Hence

$24\sum_{r=1}^{n-1}r^2=(2n-1)^3-2(n-1)-1$

$\sum_{r=1}^{n-1}r^2=\frac{(2n-1)^3-2(n-1)-1}{24}$

If we sum all the way to n, this is more commonly known as the sum of squares

$\sum_{r=1}^nr^2=\frac{n(n+1)(2n+1)}{6}$