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05 Nov 2009, 20:13
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Find n^th term of the series.
4 7 11 17 27 45...
Let me try
If we do the differences we see
differences are
3 4 6 10 18 // first level
1 2 4 8 // second level
but does this of any help ? if so, how it can help to find the general term of the series ?
4 7 11 17 27 45...
Let me try
If we do the differences we see
differences are
3 4 6 10 18 // first level
1 2 4 8 // second level
but does this of any help ? if so, how it can help to find the general term of the series ?
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05 Nov 2009, 22:51
Kudos
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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.
e.g 1st term: 3 = 2^(1-1) + 2, or
2nd term: 4 = 2^(2-1) + 2
Basically each term is of the form 2^(n-1) + 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^(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?
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.
e.g 1st term: 3 = 2^(1-1) + 2, or
2nd term: 4 = 2^(2-1) + 2
Basically each term is of the form 2^(n-1) + 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^(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?
06 Nov 2009, 08:56
Kudos
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Thanks for the response . I appreciate your hard work .
But I understood nothing
......I am even more confused than getting the concept .
Can anybody solve in simpler way ?
basically I want to derive the general formula ....I just don't know how the differences can be linked to derive the series term ?
But I understood nothing
......I am even more confused than getting the concept .Can anybody solve in simpler way ?
basically I want to derive the general formula ....I just don't know how the differences can be linked to derive the series term ?
