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)
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)
$\displaystyle \sum_{n = 0}^kar^n = a \cdot \frac{1 - r^{k + 1}}{1 - r}$
$\displaystyle \frac{1- x^{25}}{1 - x} = 10$
(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
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
$\displaystyle \sum_{n = 0}^{k}ar^n = a \cdot \frac{1 - r^{k + 1}}{1 - r}$
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