exponential function, I can't solve this problem.

**melvis** · October 11th 2009, 06:40 PM

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

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

**CFem** · October 11th 2009, 06:51 PM

An exponential function means this:

$a^x$

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.

**mr fantastic** · October 11th 2009, 06:53 PM

Get new friends

And try using $G(n) = \sum_{i=1}^{n} 2^i$ .

Originally Posted by melvis

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

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

**TheEmptySet** · October 11th 2009, 06:59 PM

Originally Posted by melvis

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

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

The pattern that you are looking for is this

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

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

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

so we get

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

This is alot of rice

**melvis** · October 12th 2009, 11:11 AM

thx for the help.

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