Finding n when (2^1)(2^2)(2^3)...(2^n) = 210

[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!

 

[Leong]

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

 

[rocketboy]

don't i feel stupid...

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!

 

[Leong]

i found an answer which is not an integer!

 

[Diane_]

i found an answer which is not an integer!

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

 

[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.