Sum of 5^1-5^2+5^3-5^4+...-5^{98}: e. (5/6)(1-5^98)

[Biosyn]

Homework Statement

Find the sum of 5^1-5^2+5^3-5^4+...-5^{98}

a. (5/4)(1-5^99)
b. (1/6)(1-5^99)
c. (6/5)(1+5^98)
d. (1-5^100)
e. (5/6)(1-5^98)

Homework Equations

The Attempt at a Solution

I feel as though this is actually a simple problem and that I'm not looking at it the right way.

[5^1 + 5^3 + 5^5...5^{97}] + [-5^2-5^4-5^6...-5^{98}]

 

[jbunniii]

Do you know how to sum $x^n$ in general? What is $x$ here?

 

[Biosyn]

Do you know how to sum $x^n$ in general? What is $x$ here?

$x$ will be 5?

\sum_{i=0}^{48} (5^{2i + 1}) + \sum_{i=0}^{49} (5^{2i})Never mind, I figured it out!

 

[jbunniii]

Actually, it looks to me like
$$-\sum_{n=1}^{98}(-5)^n$$

 

[Biosyn]

Actually, it looks to me like
$$-\sum_{n=1}^{98}(-5)^n$$

I used Sn = \frac{a_1*(1-r^n)}{1-r}

Sn = \frac{5*(1-(-5)^98)}{1-(-5)}

= \frac{5*(1-(-5)^98)}{6}

= (5/6)*(1-(-5)^98)

Thanks for your help!