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Algebra

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21 Jun 2008 11:28:46 IST

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Sum of elements

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Consider the set of numbers {1,2,3,4,5,6,7,...,n}

A number P is formed by multiplying any number of distinct elements from the set.

Find the value of $\sum \frac{1}{P}$ where the summation is carried out over all possible values of P.

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Comments (10)

Hari Shankar

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21 Jun 2008 12:27:54 IST

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A small clarification: There can be repetition in the value of P. For instance 6 = 1.6 = 2.3. But 1X6 is considered the same as 6X1. Likewise 2X3 and 3X2 are not considered as distinct for this problem

abhishek sinha

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21 Jun 2008 20:35:21 IST

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Let's try !!

Let us consider f(n ) = sum of 1/P for the set of first n natural number

then f( n+ 1 ) = sum of 1/P ( always excluding ( n+ 1 ) ) + sum of 1/P ( always taking ( n+ 1 )) + 1/ ( n+1)

= f(n ) + f( n) / ( n+1 ) + 1/( n+1)

Now solve this recursive function f(n).

abhishek sinha

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22 Jun 2008 10:04:44 IST

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Now it is very simple to solve the recursion .

Note that through actual calculation f(1) = 1 , f( 2) =2

So we propose f( i ) = i and use mathematical induction ....

so f( i+1) =i + i/( i+1) + 1/( i+1) = i+ 1( Using the relation obtained previously )

thus it is proved that f( n ) = n

so the sum is n for all natural number n .

Got it ?

Hari Shankar

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22 Jun 2008 10:45:41 IST

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There is a much more straightforward way:

You can easily see that

$\sum \frac{1}{P} = \left(1+\frac{1}{2} \right) \ \left(1+\frac{1}{3} \right) \ ...\left(1+\frac{1}{n} \right) - 1\\ \\$

$= \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times ...\times \frac{n+1}{n} - 1$

$= \frac{n+1}{2} - 1$

$= \frac{n-1}{2}$

abhishek sinha

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22 Jun 2008 12:29:08 IST

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ur soln gives for n= 1 that sum = 0 , for n= 2 , that sum = 1/2

but actually that is not so !!!!!!!!! ( 1 and 2 respectively )

so , the above solution is completely wrong !!!

Hari Shankar

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22 Jun 2008 13:13:43 IST

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Go easy on the sharp comments bro.

for n = 1, you arent multiplying any two numbers.

But even if you want to interpret it as:

$\sum \frac{1}{k} + \sum \frac{1}{1.2} ... etc.$

the sum is just $\left(1+\frac{1}{1} \right) \times \left(1+\frac{1}{2} \right) \times ...\times \left(1+\frac{1}{n} \right) -1 = n$

by the same procedure adopted above.

The focus here is on how to arrive at the answer in a manner similar to using Vieta's formulas

Hari Shankar

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22 Jun 2008 13:57:30 IST

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@Feynmann: if you can take some words of advice from someone who has seen more summers than you have, this one-upmanship in trying to score points over someone is exhausting in the long run and also a perilous enterprise. You never know when you are gonna fall flat on your face and look silly.

Instead, try to have a more mature approach of being helpful to the aspirants who are looking to follow in your footsteps. This will save you a lot of heartburn and perhaps win you some friends in the process.

abhishek sinha

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22 Jun 2008 15:31:40 IST

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Thanks for the advice ,sir !!

First of all , back to the problem

You wrote ,

"A number P is formed by multiplying ANY number of distinct elements from the set."

Now that 'ANY' may rightly be one ( can u deny that ? ), so as per the question ur first solution does not give the correct solution .

I just mean to say that and it is a completely academic comment , with no other intention . So please don't take it personally .

Secondly and most importantly , I think ( with very little experience of mine ) that it is important to make some original contribution rather than copy-pasting other's work !!!

Sir, I admire you and feel proud on the thought that I am on the same forum as that of you and my comments , solutions ( which may be just some piece of junks for someone )or whatever ..............are printed side by side of ur extremely beautiful solutions of the problems .

The solutions you provide reflect the depth of your study and ur crystal-clear thought process ( which , honestly ,I can't achieve throughout my life ) . So where the question of rivalry comes into the scene ? After all I am just a third year engineering student and knows almost nothing about the subject which is called mathematics !!

As to ur comment that , I am doing it ( calling a spade , a spade ) just because it would earn me more points or whatever , please let me remind you that these points won't cost a li'l penny ,or a single extra mark in my examination , and u can consider me mature enough who understands it !!

But , I know what is the problem with me . The problem with me is that I look at the figures on the screen rather than the name of the figure who writes it !! As a result of it , I don't hesitate to call a wrong solution a wrong one .

Again , I believe in the originalty of any solution ,however ugly it be rather than posting some extremely elegant solution of the problem which , unfortunately , has not been done by me ( in this case , Vieta,as mentioned by you) !!Please note that I am not talking of other learner's as they would surely be benefitted by seeing the same solution by two different methods ( or atleast by yours ) . It is completely my personal feelings , nothing else !!

Again sir , I feel extremely guilty on the whole process that is going on in this thread and am ready to accept whatever advice or rather punishment you give me , because I CALL YOU SIR , AND U HAVE EVERY RIGHT TO scorn me and I would accept it without hesitation as I know that it would be a blessing in the disguise .

Thanking you .

Conjurer

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22 Jun 2008 16:43:23 IST

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Calm down feynmann.You didnt need to write all those seemingly rude offtopic paras in the thread.A goiit "expert" should know this .

abhishek sinha

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22 Jun 2008 18:50:44 IST

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sorry , bro !

If I thought earlier that a simple comment " the solution is completely wrong " would arise such a scuffle then , I swear, I would never write that comment .

Anyway , I apologise to all members of the GOIIT for all these and let this thread close here .

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