Thread: 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!

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 $\displaystyle 1+2+3=(2-1)+2+(2+1)=2+2+2=3(2)$

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

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

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

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

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!=]

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?

Originally Posted by matlabnoob

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!=]

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

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

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

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

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

Telescoping

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

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

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

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

all the way to

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

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

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

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

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

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

Hence

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

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

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

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