# help me to simplify this

• Feb 27th 2008, 10:05 AM
EquinoX
help me to simplify this
$\displaystyle \sum\limits_{n = 0}^{n - 1} \left({\frac {1}{2}}\right) ^ n$
• Feb 27th 2008, 11:14 AM
wingless
I assume that $\displaystyle n-1$ on sigma is a constant and not related to the summation.

$\displaystyle \sum _{k=0}^a r^k = \frac{r^{1+a}-1}{r-1}$

So,
$\displaystyle \sum_{n = 0}^{n - 1} \left({\frac {1}{2}}\right)^n = \frac{\left(\frac{1}{2}\right)^{1+(n-1)}-1}{\frac{1}{2}-1} = 2 - 2^{1-n}$
• Feb 28th 2008, 03:02 AM
gen2dan
wel wel wel
well i think the answer to that question is 1 though im not sure
• Feb 28th 2008, 03:44 AM
mr fantastic
Quote:

Originally Posted by gen2dan
well i think the answer to that question is 1 though im not sure

If it was the infinite series $\displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^n$ then the answer would indeed be 1.

But it's not so it's not.

(Note: The very first term of the given series is $\displaystyle \left( \frac{1}{2} \right)^0 = 1$ so clearly it can't sum to 1 .......)