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The sequence s1, s2, s3,.....sn,...is such that [#permalink]

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17 Aug 2010, 12:46

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Please your help:

The sequence s1, s2, s3,.....sn,...is such that \(Sn= (1/n) - (1/(n+1))\) for all integers \(n>=1\). If k is a positive integer, is the sum of the first k terms of the sequence greater than \(9/10\)?

(1) Take K as 11. So, Sum = S1 + S2 + S3 + S4 + S5+ S6 + S7 + S8 + S9 + S10 + S11 Where, S1 = 1 - (1/2) S2 = (1/2) - (1/3) S3 = (1/3) - (1/4) ... S11 = (1/11) - (1/12)

Which implies, Sum = 1 - (1/12) /* The terms like +1/2, -1/2, +1/3, -1/3 will be added to zero. Only the first and last numbers remains */ ==> Sum = 1 - 0.0XXXX > 9/10

If you take k as 12, the SUM = 1 - (1/13) which is again > 0.9 Hence SUFFICIENT

(2) K < 19 Consider K as 2. Then the sum is = 1 - (1/2) + (1/2) - (1/3) = 1 - (1/3) which is Less than 9/10 Consider K as 11, Then the sum is greater than 9/10 /* We already proved this in (1) above */ Hence (2) is In Sufficient

(1) Take K as 11. So, Sum = S1 + S2 + S3 + S4 + S5+ S6 + S7 + S8 + S9 + S10 + S11 Where, S1 = 1 - (1/2) S2 = (1/2) - (1/3) S3 = (1/3) - (1/4) ... S11 = (1/11) - (1/12)

Which implies, Sum = 1 - (1/12) /* The terms like +1/2, -1/2, +1/3, -1/3 will be added to zero. Only the first and last numbers remains */ ==> Sum = 1 - 0.0XXXX > 9/10

If you take k as 12, the SUM = 1 - (1/13) which is again > 0.9 Hence SUFFICIENT

(2) K < 19 Consider K as 2. Then the sum is = 1 - (1/2) + (1/2) - (1/3) = 1 - (1/3) which is Less than 9/10 Consider K as 11, Then the sum is greater than 9/10 /* We already proved this in (1) above */ Hence (2) is In Sufficient

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If you notice, the sum is simply given by Sn = 1 -1/(n+1) = n/(n+1). hence S9 = 9/10. Adding further terms only increases the sum - hence S10 > S9 etc... 1) k>10, clearly sufficient as S10>9/10 = S9 2) k<19, cant say much K can be 1 = 1/2<9/10 but K=11 makes 11/12>9/10 Hence A _________________

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Example: (1) K > 10. The lowest range number here is: 11. If 11 is sufficient then take 12. If 12 is also sufficient find out if you can make a generalized statement as Sufficient?

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The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - [#permalink]

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29 Oct 2010, 20:43

The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - (1/(n+1)) for all integers n>=1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10? 1) k > 10 2) k < 19

The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - (1/(n+1)) for all integers n>=1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10? 1) k > 10 2) k < 19

Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So: \(s_1=1-\frac{1}{2}\); \(s_2=\frac{1}{2}-\frac{1}{3}\); \(s_3=\frac{1}{3}-\frac{1}{4}\); ...

If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).

Question: is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)? --> is \(\frac{k}{k+1}>\frac{9}{10}\)? --> is \(k>9\)?

(1) k > 10. Sufficient. (2) k < 19. Not sufficient.

The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - (1/(n+1)) for all integers n>=1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10? 1) k > 10 2) k < 19

I) (1/n) - (1/n+1) where N>=1; k = sum of the sequence;

Great solution by Bunuel. Good that you didn't solve the sn formula initially to 1/n(n+1) . that way its easy to cancel out terms.

Bunuel wrote:

satishreddy wrote:

The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - (1/(n+1)) for all integers n>=1. If k is a positive integer, is the sum of the first k terms of the sequence greater than 9/10? 1) k > 10 2) k < 19

Given: \(s_n=\frac{1}{n}-\frac{1}{n+1}\) for \(n\geq{1}\). So: \(s_1=1-\frac{1}{2}\); \(s_2=\frac{1}{2}-\frac{1}{3}\); \(s_3=\frac{1}{3}-\frac{1}{4}\); ...

If you sum the above 3 terms you'll get: \(s_1+s_2+s_3=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})=1-\frac{1}{4}\) (everything but the first and the last numbers will cancel out). So the sum of first \(k\) terms is fgiven by the formula \(sum_k=1-\frac{1}{k+1}\).

Question: is \(sum_k=1-\frac{1}{k+1}>\frac{9}{10}\)? --> is \(\frac{k}{k+1}>\frac{9}{10}\)? --> is \(k>9\)?

(1) k > 10. Sufficient. (2) k < 19. Not sufficient.

If you notice, the sum is simply given by Sn = 1 -1/(n+1) = n/(n+1). hence S9 = 9/10. Adding further terms only increases the sum - hence S10 > S9 etc... 1) k>10, clearly sufficient as S10>9/10 = S9 2) k<19, cant say much K can be 1 = 1/2<9/10 but K=11 makes 11/12>9/10 Hence A

I don't think sn=1/n-1/(n+1) is equal to n/(n+1), it's equal to 1/n(n+1).

I find that sequence problems come down to pattern recognition, nravi's solution is probably what it's supposed to look like, you have to see that the terms other than 1 and 1/12 cancel each other out.

If you notice, the sum is simply given by Sn = 1 -1/(n+1) = n/(n+1). hence S9 = 9/10. Adding further terms only increases the sum - hence S10 > S9 etc... 1) k>10, clearly sufficient as S10>9/10 = S9 2) k<19, cant say much K can be 1 = 1/2<9/10 but K=11 makes 11/12>9/10 Hence A

I don't think sn=1/n-1/(n+1) is equal to n/(n+1), it's equal to 1/n(n+1).

I find that sequence problems come down to pattern recognition, nravi's solution is probably what it's supposed to look like, you have to see that the terms other than 1 and 1/12 cancel each other out.

sum of \([\frac{1}{n} - \frac{1}{(n+1)}]\) from = 1 to n=n => \(\frac{n}{(n+1)}\) \(1-\frac{1}{2} + \frac{1}{2}-\frac{1}{3} + \frac{1}{3}-\frac{1}{4} + ............... \frac{1}{n} -\frac{1}{(n+1)}\) = \(1 - \frac{1}{(n+1)}\) =\(\frac{n}{(n+1)}\) because all the terms between 1st and last are cancelled. _________________

If you notice, the sum is simply given by Sn = 1 -1/(n+1) = n/(n+1). hence S9 = 9/10. Adding further terms only increases the sum - hence S10 > S9 etc... 1) k>10, clearly sufficient as S10>9/10 = S9 2) k<19, cant say much K can be 1 = 1/2<9/10 but K=11 makes 11/12>9/10 Hence A

I don't think sn=1/n-1/(n+1) is equal to n/(n+1), it's equal to 1/n(n+1).

I find that sequence problems come down to pattern recognition, nravi's solution is probably what it's supposed to look like, you have to see that the terms other than 1 and 1/12 cancel each other out.

sum of \([\frac{1}{n} - \frac{1}{(n+1)}]\) from = 1 to n=n => \(\frac{n}{(n+1)}\) \(1-\frac{1}{2} + \frac{1}{2}-\frac{1}{3} + \frac{1}{3}-\frac{1}{4} + ............... \frac{1}{n} -\frac{1}{(n+1)}\) = \(1 - \frac{1}{(n+1)}\) =\(\frac{n}{(n+1)}\) because all the terms between 1st and last are cancelled.

Re: The sequence s1, s2, s3,.....sn,...is such that [#permalink]

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07 Jan 2012, 16:08

Tricky one, needs so serious rephrasing of the stem:

Rephrase: for the k number of elements, the sum of all the terms will be = 1/x - 1/(x+k) (try yourself! - intermediate terms cancel out)

1. Say k = 11, we get 1/1 - 1/(1+11) = 11/12 > 9/10. Suff 2. k<19. it could be k=11 as above or k=8, where you get 1/1 - 1/(1+8) = 8/9 < 9/10. Insuff.

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Re: The sequence s1, s2, s3,.....sn,...is such that Sn= (1/n) - [#permalink]

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25 Dec 2013, 11:03

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