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Re: general term of the series [#permalink]
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05 Nov 2009, 22:51
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^(n1) 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.
e.g 1st term: 3 = 2^(11) + 2, or 2nd term: 4 = 2^(21) + 2
Basically each term is of the form 2^(n1) + 2
(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^(n1) + 2. Note we need to adjust this to 2^(n2) + 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^(22) + 2 = =4 + 1+ 2 = 7  it works! 3rd term = 2nd term + 2^(32) + 2 = 7 + 2 + 2 = 11  it works! ..and so on.
nth term = (n1)th term + 2^(n2) + 2 A (n1)th term = (n2)th term + 2^(n3) + 2 B
Applying B in A, we get nth term = (n2)th term + 2^(n3) + 2^(n2) + 2 + 2
Clearly, we can put in the formula for the (n2)th term in terms of the (n3)th term and so on till we reach the first term 4.
That would look like: nth term = 1st term + [2^(11) + 2^(21) + 2^(31)+...+2^(n2)] + 2*(n1) = 4 + 2*(n1) + [1+2+4+...+2^(n2)] = 4 + 2*(n1) + [(2^(n1)  1) / (21)] using the formula for a geometric progression = 4 + 2*(n1) + 2^(n1)  1 = 4 + 2*n 2 + 2^(n1)  1 = 1 + 2*n + 2^(n1)
So the nth term of the series = 1 + 2*n + 2^(n1) 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?
