sum of the series
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tomada
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I calculated the answer to be 1/4. I don't think you'll see this type of question on the GMAT, though.
I'm really old, but I'll never be too old to become more educated.
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tomada
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First I noticed that 1/8 can be factored out of each term, as follows:
1/8 * (1/1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + ...)
Notice an interesting pattern with the denominators inside the parentheses?
From the first term to the second term, the difference between the denominators = 2
From the second term to the third term, the difference between the denominators = 3
From the third term to the fourth term, the difference between the denominators = 4
..and so on
It took some trial-and-error on my part, but the terms within the parentheses can be expressed as the sum of all terms given by 1/((n^2 + n)/2), as 'n' goes from 1 to infinity.
For example, when n=1, 1/((1^2 + 1)/2) = 1/1
...when n=2, 1/((2^2 + 2)/2) = 1/3
...when n=3, 1/((3^2 + 3)/2) = 1/6
...when n=4, 1/((4^2 + 4)/2) = 1/10, etc.
Also, note that 1/((n^2 + n)/2) can be expressed as 2/(n^2 + n)
Now, the entire sum = 1/8 * (sum of all terms given by 2/(n^2 + n), as 'n' goes from 1 to infinity)
The '2' in the numerator can be factored out as well, resulting in:
1/4 * (sum of all terms given by 1/(n^2 + n), as 'n' goes from 1 to infinity)
However, the expression 1/(n^2 + n) = 1/((n)(n+1)) = 1/n - 1/(n+1)
When expressed as 1/n - 1/(n+1), the terms become (1-1/2) + (1/2 - 1/3) + (1/3 - 1/4), etc...
In the end, all the terms are cancelled out, except for the '1' at the beginning.
So, the overall sum is now 1/4 (which was the constant taken outside the series) * 1 = 1/4.
1/8 * (1/1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + ...)
Notice an interesting pattern with the denominators inside the parentheses?
From the first term to the second term, the difference between the denominators = 2
From the second term to the third term, the difference between the denominators = 3
From the third term to the fourth term, the difference between the denominators = 4
..and so on
It took some trial-and-error on my part, but the terms within the parentheses can be expressed as the sum of all terms given by 1/((n^2 + n)/2), as 'n' goes from 1 to infinity.
For example, when n=1, 1/((1^2 + 1)/2) = 1/1
...when n=2, 1/((2^2 + 2)/2) = 1/3
...when n=3, 1/((3^2 + 3)/2) = 1/6
...when n=4, 1/((4^2 + 4)/2) = 1/10, etc.
Also, note that 1/((n^2 + n)/2) can be expressed as 2/(n^2 + n)
Now, the entire sum = 1/8 * (sum of all terms given by 2/(n^2 + n), as 'n' goes from 1 to infinity)
The '2' in the numerator can be factored out as well, resulting in:
1/4 * (sum of all terms given by 1/(n^2 + n), as 'n' goes from 1 to infinity)
However, the expression 1/(n^2 + n) = 1/((n)(n+1)) = 1/n - 1/(n+1)
When expressed as 1/n - 1/(n+1), the terms become (1-1/2) + (1/2 - 1/3) + (1/3 - 1/4), etc...
In the end, all the terms are cancelled out, except for the '1' at the beginning.
So, the overall sum is now 1/4 (which was the constant taken outside the series) * 1 = 1/4.
I'm really old, but I'll never be too old to become more educated.
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aneesh.kg
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There is certain procedure to solve such a problem. And, the method is difficult to occur to you unless you've been explained the method atleast once.
So, here we go.
Read carefully, and write down the steps as you read it.
1/2 - 1/4
(Taking the common denominator to be the product of 2 and 4)
= (4 - 2)/(2*4) = 2/(2*4)
so
1/2*4 = (1/2)*(1/2 - 1/4)
Each term in the series can be written in the same manner.
1/4*6 = (1/2)*(1/4 - 1/6)
..
1/18*20 = (1/2)*(1/18 - 1/20)
The sequence is
1/(2*4) + 1/(4*6) +1/(6*8) + 1/(8*10).......... +1/(18*20)
(Substituting the value of each of the above terms from above in this problem)
=
(1/2)*(1/2 - 1/4) + (1/2)*(1/4 - 1/6).... (1/2)*(1/16 - 1/18) +(1/2)*(1/18 - 1/20)
(Taking 1/2 common from each term)
=
(1/2)*[(1/2 - 1/4) + (1/4 - 1/6).... (1/16 - 1/18) +(1/18 - 1/20)]
(The eyes should lighten up now because all terms except 1/2 and 1/20 cancel off with each other)
=
(1/2)*[1/2 - 1/20]
=
(1/2)*[9/20]
=
[spoiler]9/40[/spoiler]
Practice Problem:
Find the value of
(1/(3*6) + 1/(6*9) + .............. 1/(33*36)]
So, here we go.
Read carefully, and write down the steps as you read it.
1/2 - 1/4
(Taking the common denominator to be the product of 2 and 4)
= (4 - 2)/(2*4) = 2/(2*4)
so
1/2*4 = (1/2)*(1/2 - 1/4)
Each term in the series can be written in the same manner.
1/4*6 = (1/2)*(1/4 - 1/6)
..
1/18*20 = (1/2)*(1/18 - 1/20)
The sequence is
1/(2*4) + 1/(4*6) +1/(6*8) + 1/(8*10).......... +1/(18*20)
(Substituting the value of each of the above terms from above in this problem)
=
(1/2)*(1/2 - 1/4) + (1/2)*(1/4 - 1/6).... (1/2)*(1/16 - 1/18) +(1/2)*(1/18 - 1/20)
(Taking 1/2 common from each term)
=
(1/2)*[(1/2 - 1/4) + (1/4 - 1/6).... (1/16 - 1/18) +(1/18 - 1/20)]
(The eyes should lighten up now because all terms except 1/2 and 1/20 cancel off with each other)
=
(1/2)*[1/2 - 1/20]
=
(1/2)*[9/20]
=
[spoiler]9/40[/spoiler]
Practice Problem:
Find the value of
(1/(3*6) + 1/(6*9) + .............. 1/(33*36)]
Aneesh Bangia
GMAT Math Coach
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GMAT Math Coach
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GMATPad:
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tomada
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In Aneesh' solution, the sum of infinite terms = 9/40, which equals 0.225.
The first nine terms of the series are 1/8, 1/24, 1/48, 1/80, 1/120, 1/168, 1/224, 1/288, 1/360.
These terms add up to 0.225, so any terms thereafter will make the sum greater than 9/40.
Therefore, the answer cannot be 9/40.
The first nine terms of the series are 1/8, 1/24, 1/48, 1/80, 1/120, 1/168, 1/224, 1/288, 1/360.
These terms add up to 0.225, so any terms thereafter will make the sum greater than 9/40.
Therefore, the answer cannot be 9/40.
I'm really old, but I'll never be too old to become more educated.
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tomada
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Oops, I goofed. I thought this was an infinite series, but now I realize that the series ends at 1/(18*20), which means that Aneesh' solution is correct.
I solved for an infinite series, meaning that the next terms would have been 1/(20*22) + 1/(22*24) + 1/(24*26)....
On another note, how does one delete his own posts? All I see are the "Quote" and "Edit" buttons.
I solved for an infinite series, meaning that the next terms would have been 1/(20*22) + 1/(22*24) + 1/(24*26)....
On another note, how does one delete his own posts? All I see are the "Quote" and "Edit" buttons.
I'm really old, but I'll never be too old to become more educated.
