I've been struggling with this and I even asked 2 friends of mine to help me but they couldn't solve it.

http://www.mathhelpforum.com/math-he...1&d=1255315137

Unfortunately, I can't find the function :Z. I hope somebody can help me. thx 4 u time.

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- Oct 11th 2009, 06:40 PMmelvisexponential function, I can't solve this problem.
I've been struggling with this and I even asked 2 friends of mine to help me but they couldn't solve it.

http://www.mathhelpforum.com/math-he...1&d=1255315137

Unfortunately, I can't find the function :Z. I hope somebody can help me. thx 4 u time. - Oct 11th 2009, 06:51 PMCFem
An exponential function means this:

$\displaystyle a^x$

If you notice, from 1->2 and 2->3 the value doubles. That's not always a valid hint, but it works for this problem.

So use 2 as the base and see if you can figure out the exponent. - Oct 11th 2009, 06:53 PMmr fantastic
- Oct 11th 2009, 06:59 PMTheEmptySet

The pattern that you are looking for is this

$\displaystyle \{2,4,8,16,.... \}=\{2^1,2^2,2^3,...,2^{64} \}$.

So write this as a sum you get $\displaystyle \sum_{n=1}^{64}2^n$

$\displaystyle \sum_{n=0}^{64}2^n=1+\sum_{n=1}^{64}2^n=\frac{1-2^{64+1}}{1-2}=2^{65}-1$

so we get

$\displaystyle 1+\sum_{n=1}^{64}2^n=2^{65}-1 \iff \sum_{n=1}^{64}2^n=2^{65}-2=36893488147419103230$

This is alot of rice - Oct 12th 2009, 11:11 AMmelvis
thx for the help.