# Stuck on review math problem

1. Sep 20, 2005

### rocketboy

Hey everyone, I am stuck on this question:
$$(2^1)(2^2)(2^3)(2^4)...(2^n) = 210$$

What is the value of n?

I'd love to post what I have so far, except that the problem doesn't involve much work, so as soon as I have the first step I'll have the last...and I haven't yet gotten the first (correct) step.

THANKS!!

Last edited: Sep 20, 2005
2. Sep 20, 2005

### Leong

$$2^{1+2+3+...+n}$$=210

3. Sep 20, 2005

### rocketboy

don't i feel stupid.... :rofl:

I was doing the exact same thing only with the tn = a + (n-1)d formula instead of the Sn = n/2[2a + (n-1)d] formula!!!

Thank you!

Last edited: Sep 20, 2005
4. Sep 20, 2005

### Leong

i found an answer which is not an integer!

5. Sep 20, 2005

### Diane_

Since 210 is not a multiple of 2, that's to be expected.

6. Sep 20, 2005

### Leong

it is contrary to the method, because 1 + 2+ 3 +.. +n indicates that n must be an integer if want to use the arithmetic sequence formulae.