# Geometric series

• May 7th 2011, 08:52 AM
elieh
Geometric series
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?
• May 7th 2011, 09:13 AM
HallsofIvy
The previous LaTeX routine, that used "math" and "/math" isn't working. However, it appears that using "tex" and "\tex" will work:
\$\displaystyle r^3+ r^2+ r- 14= 0\$

To show that "r= 2 is the only real solution for this equation", first show that r= 2 is a solution:
\$\displaystyle 2^3+ 2^2+ 2- 14= 8+ 4+ 2- 14= 0\$

And since r= 1 is soluton, x- 2 is a factor of that equation:
\$\displaystyle r^3+ r^2+ r- 14= (x- 2)(ax^2+ bx+ c)\$

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 \$\displaystyle r^3+ r^2+ r- 14\$ by x- 2- and showing that the quadratic has no real zeros.
• May 7th 2011, 09:14 AM
Plato
Quote:

Originally Posted by elieh
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.

Use the [tex] tags [tex](2)^2+(2)^2+(2)-14=0[/tex] gives \$\displaystyle (2)^2+(2)^2+(2)-14=0\$.

Note that \$\displaystyle r^3+r^2+r-14=(r-2)(r^2+3r+7)\$
• May 7th 2011, 10:36 AM
elieh
Makes sense,
Thanks guys.
Test:
\$\displaystyle x^2\$