# Math Help - Geometric series

1. ## Geometric series

Find the sum to n terms of this series:

1 -y +y^2 -y^3 ...

I have figured that a=(-1)^n+1.
r=-y.

2. You are correct in saying that r=-y, but you're overcomplicating the other part: a is just the first term in this series. In this case, a = 1. Now you just need to substitute the values into the formula.

3. That works but it isn't the answer in the book.

It is:

1+((-1)^n+1) * y^n
divided by 1+y.

4. Hmm...have you posted the question correctly? I can't see any error in my logic.

5. You and the book are correct. The answer in the book confused me. They must have decided to take the answer a step further.

6. You are considering $r=-y$. And the first term of your series is :a= 1.

so the sum of the geometric series would be given by: $S = \frac{a(1-{r^n})}{1-r}$

try plugging in the values now

7. I've already done that after the first post by Quacky.

8. I think the textbook is either misprinted or has found some other values which also work, but I don't believe my calculation is incorrect.

9. Originally Posted by Stuck Man
Find the sum to n terms of this series:

1 -y +y^2 -y^3 ...

I have figured that a=(-1)^n+1.
r=-y.
$a=1$

$r=-y$

$S_n=\displaystyle\frac{a\left(1-r^n\right)}{1-r}=\frac{a\left(r^n-1\right)}{r-1}$

$=\displaystyle\frac{1-r^n}{1-r}=\frac{r^n-1}{r-1}$

$=\displaystyle\frac{1-(-y)^n}{1-(-y)}=\frac{(-y)^n-1}{-y-1}$

$=\displaystyle\frac{1-(-1)^ny^n}{1+y}=\frac{(-1)^ny^n-1}{-(y+1)}$

$=\displaystyle\frac{1-(-1)^ny^n}{1+y}=\frac{1+(-1)(-1)^ny^n}{1+y}=\frac{1+(-1)^{n+1}y^n}{1+y}$

10. Thats more work than necessary. You don't need to have loads of steps to produce the coefficient of -1 or 1.

11. Originally Posted by Stuck Man
Thats more work than necessary. You don't need to have loads of steps to produce the coefficient of -1 or 1.
The steps are only for clarity.