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The idea of doing the differences makes sense. Haven't done anything like this since high school so I'm really out of touch - took me loads of time to figure this out.

First level difference is 3,4,6,10,18 Second level difference is 1,2,4,8 ----> this is basically powers of 2 i.e. 2^(n-1) for n=1,2,...

To connect the first level and second level difference, we can see that each number in the 1st level (3,4,6,10,18..) is just the 2nd level number (1,2,4,8,..) plus a 2 added to it.

(I guess it's ideal if you just look at the first level difference of 3,4,6,10,18.. and just see that you can get each number by adding 2 to consecutive powers of 2)

So now we have a formula for the differences between the terms of the original given series. The difference between 2 consecutive terms is of nature 2^(n-1) + 2. Note we need to adjust this to 2^(n-2) + 2 as the first difference corresponds to the 2nd term in the original series.

The original given series is 4,7,11,17,27,45...

So 2nd term = 1st term + 2^(2-2) + 2 = =4 + 1+ 2 = 7 - it works! 3rd term = 2nd term + 2^(3-2) + 2 = 7 + 2 + 2 = 11 - it works! ..and so on.

nth term = (n-1)th term + 2^(n-2) + 2 -----------A (n-1)th term = (n-2)th term + 2^(n-3) + 2 -----------B

Applying B in A, we get nth term = (n-2)th term + 2^(n-3) + 2^(n-2) + 2 + 2

Clearly, we can put in the formula for the (n-2)th term in terms of the (n-3)th term and so on till we reach the first term 4.

That would look like: nth term = 1st term + [2^(1-1) + 2^(2-1) + 2^(3-1)+...+2^(n-2)] + 2*(n-1) = 4 + 2*(n-1) + [1+2+4+...+2^(n-2)] = 4 + 2*(n-1) + [(2^(n-1) - 1) / (2-1)] -----using the formula for a geometric progression = 4 + 2*(n-1) + 2^(n-1) - 1 = 4 + 2*n -2 + 2^(n-1) - 1 = 1 + 2*n + 2^(n-1)

So the nth term of the series = 1 + 2*n + 2^(n-1) Phew! I know I'm rusty and this took me quite some time. But I don't see how this can be a GMAT question if it's to be done in 2 mins - is there a faster way?