A geometric series has first term a and common ratio r where r\geqslant 1
The sum of the first four terms of this series is 15 times the first term.
a) show that r^3 + r^2 +r -14=0
and having S4 = 15a
I've done the first part
But I don't get b
b) Show that r=2 is the only real solution for this equation.
Every time I use Latex I get an error when I preview, has anything changed?
The previous LaTeX routine, that used "math" and "/math" isn't working. However, it appears that using "tex" and "\tex" will work:
To show that "r= 2 is the only real solution for this equation", first show that r= 2 is a solution:
And since r= 1 is soluton, x- 2 is a factor of that equation:
What you need to do is find that quadratic factor- find a, b, c by, perhaps by multiplying out the right side and setting corresponding cooefficients equal, getting three equations for a, b, and c, perhaps by dividing by x- 2- and showing that the quadratic has no real zeros.
Use the [tex] tags [tex](2)^2+(2)^2+(2)-14=0[/tex] gives .
Originally Posted by elieh