## GMAT Advanced Quant: Try-It #0-4

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StefanS348
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Posts: 1
Joined: Sat Aug 06, 2016 6:21 am

### GMAT Advanced Quant: Try-It #0-4

Hi! When I completed this question I used an alternate pattern which appears to work too.

What should the next number in this sequence be?

1 2 9 64 ??

The way I answered it was as follows:

n sub q = the value of a a given term in the sequence (i.e. n sub q=1, 2, 9,...., ??, depending on q. Sorry for the weird notation, but I'm not sure how to make subscripts on my phone so I have to use sub q)

q = term of the sequence (q=1, 2, 3, 4.... So when q=3, n=9 or in other words, n sub (q=3) = 9)

You can find any number in the sequence by using this formula:

n sub (q+1) = (n sub q)(1+3(q-1))+1

n sub (1+1) = (1)(1 + 0) + 1 = 2
n sub (2 + 1) = (2)(1 + 3(2 - 1)) + 1 = 9
n sub (3 + 1) = (9)(1 + 3(3 - 1)) + 1 = 64
n sub (4 + 1) = (64)(1 + 3(4 - 1)) + 1 = 641

This can also be used to work backwards if, for example, you weren't given that n sub 1 = 1.

Thanks for your help and I look forward to hearing back!
RonPurewal
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Posts: 19746
Joined: Tue Aug 14, 2007 8:23 am

### Re: GMAT Advanced Quant: Try-It #0-4

well, ok, so... first of all you can rest easy -- the actual test will NEVER give you a problem that's this sketchy and ill-defined. (on the contrary, the official test specifies stuff so exactly that it's often annoying... literally to the extent of providing definitions for things that are already perfectly obvious. but, of course it's better to be too specific than too vague.)

that said...
you have a formula there with at least 3 degrees of freedom. theoretically, i think you could apply that sort of formula to ANY sequence of 3 or 4 initial values and come up with something (like a curve fitting problem in statistics).
if you've done enough problems for this exam then you'll know this already, but, this is TOTALLY not the kind of "pattern" you'll ever have to discern on this exam!

the patterns on this exam are, over and above absolutely all else, SIMPLE patterns. if you are actually INVENTING a formula that looks like the one you have there -- as opposed to being GIVEN that formula, and just finding the unknowns (which would be more legitimate) -- then... nooooo waaaayyyyy.

--

what i'm guessing they have in mind here is...
1^0
2^1
3^2
4^3

...and so the next number in the list would be 5^4 = 625?
what does the answer key show?

in any case, the numbers 9 and 64 should be immediately recognizable as POWERS of smaller integers. from there you could work out the pattern above, even if it didn't occur to you immediately. (it didn't occur to me immediately.)