so it was

If 1+x+x^2+x^3+.....+x^24=10 what is x

so i got

((x^25) - x) / (x -1) = 9

and i put in my calc and got the right answer..but how can i do it just by solving the algebra?

the answer is .90927693 (almost positive)

- Mar 9th 2008, 10:00 AMstones44Geometric Series....got it down to algebra...but i can figure it out
so it was

If 1+x+x^2+x^3+.....+x^24=10 what is x

so i got

((x^25) - x) / (x -1) = 9

and i put in my calc and got the right answer..but how can i do it just by solving the algebra?

the answer is .90927693 (almost positive) - Mar 9th 2008, 10:10 AMstones44another one
also

1+x+x^2+x^3........=10

set it up: (0-x)/(x-1) = 9 so x = .9

but i did it using my calc also

im clearly missing a big piece of algebra knowledge - Mar 9th 2008, 10:42 AMtopsquark
$\displaystyle \sum_{n = 0}^kar^n = a \cdot \frac{1 - r^{k + 1}}{1 - r}$

$\displaystyle \frac{1- x^{25}}{1 - x} = 10$

(Note the difference between my formula and yours. I took a0 = 1 and n = 24. You took a0 = x and n = 24. Clearly your n should have been 23.)

So:

$\displaystyle 1 - x^{25} = 10(1 - x)$

$\displaystyle 1 - x^{25} = 10 - 10x$

$\displaystyle x^{25} - 10x + 9 = 0$

You want to find an exact root of this equation? Can't be done in general. There are no rational roots, at least, so we are stuck with numerical approximation.

-Dan - Mar 9th 2008, 10:45 AMtopsquark
This one, ironically, we can solve.

$\displaystyle \sum_{n = 0}^{\infty}ar^n = \frac{a}{1 - x}$

So

$\displaystyle 1 + x + x^2 + ~...~ = \frac{1}{1 - x} = 10$

$\displaystyle 1 = 10(1 - x)$

$\displaystyle 1 = 10 - 10x$

$\displaystyle 10x = 9$

$\displaystyle x = \frac{9}{10}$

-Dan - Mar 9th 2008, 10:46 AMstones44
arent you supposed to do last *multipler - first / multipler - 1

and the multipler is X and first is x because 1 is like a seperate thing (not part of the series) - Mar 9th 2008, 11:02 AMtopsquark
$\displaystyle \sum_{n = 0}^{k}ar^n = a \cdot \frac{1 - r^{k + 1}}{1 - r}$

Here you have

$\displaystyle \sum_{n = 0}^{24} 1 \cdot x^k = 1 + x + x^2 + ~ ... ~ + x^{23} + x^{24}$

So a = 1, the coefficient of $\displaystyle x^k$ in the sum, otherwise known as the first term in the summation, and r = x, the geometric ratio.

What you tried to do was this:

$\displaystyle 1 + (x + x^2 + ~...~ + x^{23} + x^{24} ) = 10$

$\displaystyle x + x^2 + ~...~ + x^{23} + x^{24} = 9$

There is nothing wrong in doing this, but now your series is

$\displaystyle \sum_{n = 0}^{23}x \cdot x^k = x \cdot \frac{1 - x^{23}}{1 - x}$

so a = x and r = x.

-Dan