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Cambridge tests -plz explain

by arora007 » Thu Jan 20, 2011 6:06 am
What is the 999th term of the series S?
1) the first 4 terms of S are (1+1)^2 , (2+1)^2, (3+1)^2, (4+1)^2
2) for every x, the xth term of S is (x+1)^2

[spoiler]
I chose D, however the OA is given as B.
OE says (1) looks good, but does not actually tell us that the series continues in the same manner beyond the terms listed' thus it is not sufficient.
[/spoiler]

I am still not able to digest, why?
is this not a borderline question?
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by advita » Thu Jan 20, 2011 6:26 am
i chose B.
for 1st...dont give any idea for after 4th terms.... may it it is cyclic from 5th.
for 2, for xth gterm i.e x is integers so sufficient.

B.
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by Rahul@gurome » Thu Jan 20, 2011 6:59 am
arora007 wrote:I chose D, however the OA is given as B.
OE says (1) looks good, but does not actually tell us that the series continues in the same manner beyond the terms listed' thus it is not sufficient.
Statement 1 tells us specifically about the first four terms, which is not sufficient to conclude that all terms of the series also follows that particular trend. It may be possible that the terms after the 4th term doesn't follow it. Hence NOT sufficient.

Whereas statement 2 clearly states all terms are of that particular form. Hence SUFFICIENT.

The correct answer is B.
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by arora007 » Thu Jan 20, 2011 9:15 am
can the number 1 , 2, 3, 4... in a series be expressed in such a manner that they are not following that trend.

I recall a sketchy theorem from school..

if P(1) = true , and if P(n) = true and P(n+1) = true then for all n belonging to natural numbers P(n) is true .

"Don't exactly recall this theorm - if one could quote"

here

P(1) P(3) P(3+1) are true then for all n natural numbers this is true. isn't it ?
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by Rahul@gurome » Thu Jan 20, 2011 9:34 am
arora007 wrote:can the number 1 , 2, 3, 4... in a series be expressed in such a manner that they are not following that trend.
First of all, the term "series" always does not mean that there must exist any particular relation between the terms. Any sequence of integers or number can be a series. So unless we are provided with a particular relation or some relations independent of the terms, we cannot assume any trend.

Also if you are so determined that the series 4, 9, 16, 25, ... follows only a particular trend, i.e. (term number + 1)², I can give you some counterexamples. The series may be defined as any of the following
  • 1. x-th term = (x+1)² for x is not divisible by 5 and x-th term = K (some constant), if x is divisible by 5
    2. x-th term = (x+1)² for x < 5 and x-th term = x² , if x ≥ 5
    3. x-th term = (x+1)² for x ≤ 10 and x-th term = x , if x > 10
    4. ...
    5. ...
I can give you 100 such examples.
arora007 wrote:I recall a sketchy theorem from school..

if P(1) = true , and if P(n) = true and P(n+1) = true then for all n belonging to natural numbers P(n) is true.
There is no such "theorem".
I believe you are trying to mention a particular method known as Mathematical Induction, which is a higher level algebraic concept. I can go into the details of it and show you where you're misunderstanding it, but it's not really necessary.
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by arora007 » Thu Jan 20, 2011 9:50 am
Also if you are so determined
I can give you 100 such examples.
Chill... no offenses from my end!!

btw thanks for the "Mathematical Induction" and the overall explanation.

If P(0) holds and
whenever P(k) is true then P(k+1) is also true
then P(n) holds for all n.

The principle correctly proves that the terms 1,2,3,4 in (1+1)^2 , (2+1)^2, (3+1)^2, (4+1)^2... cannot be part of a series as P(0) does not hold.
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by Adam@Knewton » Thu Jan 20, 2011 2:25 pm
arora007 wrote: If P(0) holds and
whenever P(k) is true then P(k+1) is also true
then P(n) holds for all n.

The principle correctly proves that the terms 1,2,3,4 in (1+1)^2 , (2+1)^2, (3+1)^2, (4+1)^2... cannot be part of a series as P(0) does not hold.
Mathematical Induction uses P(1), not P(0), so this sequence could, in fact, be a proper series (we don't need to have a 0th term). However, Induction requires us to have some formula for which to plug in p(k+1) and also p(1) and deduce that the latter is valid and the former is valid so long as we assume p(k) true, thus proving it true for all positive integers. Statement (1) does not have a formula, which is part of the reason it's Insufficient, after all.

Incidentally, while we're on the subject of sequences of numbers with no mathematical pattern, here is my favorite -- there IS a pattern, but you have to think outside the box to get it. Any guesses? If you can post the next term, I'll be impressed.

1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221, ...
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by anshumishra » Thu Jan 20, 2011 6:34 pm
AdamKnewton wrote:
Incidentally, while we're on the subject of sequences of numbers with no mathematical pattern, here is my favorite -- there IS a pattern, but you have to think outside the box to get it. Any guesses? If you can post the next term, I'll be impressed.

1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221, ...
1321131112311311221

Here is how :-)

31131211131221
13
31131211131221
21
31131211131221
13
31131211131221
11
31131211131221
12
31131211131221
31
31131211131221
13
31131211131221
11
31131211131221
22
31131211131221
11
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by Adam@Knewton » Fri Jan 21, 2011 9:30 am
Very nice =)
anshumishra wrote:
AdamKnewton wrote:
Incidentally, while we're on the subject of sequences of numbers with no mathematical pattern, here is my favorite -- there IS a pattern, but you have to think outside the box to get it. Any guesses? If you can post the next term, I'll be impressed.

1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221, ...
1321131112311311221

Here is how :-)

31131211131221
13
31131211131221
21
31131211131221
13
31131211131221
11
31131211131221
12
31131211131221
31
31131211131221
13
31131211131221
11
31131211131221
22
31131211131221
11
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