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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Dec 2007 13:43:06 IST
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am I right?
pls somebody reply..........
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my future signature:
LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.
Happiness can be found, even in the darkest of times, if one only remembers to turn on the light.
Albus Dumbledore
Harry Potter and the Prisoner of Azkaban movie
There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.
-- Albert Einstein
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23 Dec 2007 13:54:23 IST
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NO yaar acc to the above question the ans given by him is 100% correct..
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23 Dec 2007 14:01:08 IST
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can u explain???????? why series can't go beyond Sn
if it is not mentioned in the question?
and pls give d name of book from which this question has been taken.....
I dun want to prove anyone's answer wrong but just want to confirm which answer is right......
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my future signature:
LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.
Happiness can be found, even in the darkest of times, if one only remembers to turn on the light.
Albus Dumbledore
Harry Potter and the Prisoner of Azkaban movie
There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.
-- Albert Einstein
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23 Dec 2007 14:11:48 IST
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yaar in the question there is nthn like Sn+1 which has to mentioned properly.. As Sn+1 is nt given as it is in GP so assume it to be sme general term
i dunno frm wer it is taken!!!
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23 Dec 2007 14:13:00 IST
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Here's ma solution:
The Sum of the series
a + ar + ar2 + ........ = a/1-r
Fom this the sum os the given series = Sk
Sk = k/ { 1 - (1/k+1) }
Sk = k+1
So,
S1 = 1+1 = 2
Now the sum of the series.
S12+S22+S32+...............
= 22 + 32 + ....... (2n)2
=12 + 22 + 32 + ........ + (2n)2 - 12
(Adding & Substracting 12)
We know for series of "n" terms of 12+22+.......+(n)2
S = n(n+1)(2n+1)/6
So appplying the same for "2n" terms we have:
S12+S22+S32+...............
= [ (2n)(2n+1)(4n+1)/6 ] - 1
= [ (2n)(2n+1)(4n+1)-6/6 ]
= (n)(2n+1)(4n+1)-3
3
Hope you all find it useful.
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Cheers!!!!!!!!!!!!!!!
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Always available for help !
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23 Dec 2007 14:14:58 IST
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@ waterdemon
read the whole forum first....
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my future signature:
LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.
Happiness can be found, even in the darkest of times, if one only remembers to turn on the light.
Albus Dumbledore
Harry Potter and the Prisoner of Azkaban movie
There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.
-- Albert Einstein
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23 Dec 2007 14:55:18 IST
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oh sry lokesh, didn't read ur soln properly. u were close definitely. and cHimanshu was right. only smthing waz to be added to ur answer.
S1, S2, S3, ...., Sn are clearly mentioned sums. question doesn't say anything beyond dat. if it had been given dat Sn=n+1, den u cud have taken S(2n-1) = 2(n+1)-1.
Hope u got it.
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23 Dec 2007 15:07:40 IST
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see ....if it is given that n is a constant..........then ur solution is correct...
and if n is variable then Karthik's and waterdemon's solution r correct........
this is my final conclusion.. and I think if it is not mentioned,then we'll treat it as variable..
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my future signature:
LOKESH SARDANA,
department of Production and Industrial engineering,
Indian Institute of Technology,Roorkee.
Happiness can be found, even in the darkest of times, if one only remembers to turn on the light.
Albus Dumbledore
Harry Potter and the Prisoner of Azkaban movie
There are only two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle.
-- Albert Einstein
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23 Dec 2007 15:14:29 IST
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understand the language of question yaar
"If S1, S2, S3, ...., Sn are sums of infinite geometric series whose first terms are 1, 2, 3, ..., n and whose common ratios are 1/2, 1/3, 1/4, ...., 1/n+1 respectively....". Series stops at Sn. Its not like S1, S2, S3,...., Sn, ...., S(2n-1),.....and so on.
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<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
I am only one,
But still I am one.
I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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23 Dec 2007 15:30:08 IST
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answer is n(n+1)(2n+1)-6/6
sure
its correct
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23 Dec 2007 15:33:21 IST
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nopes
ans is
[ (n+1)(n+2)(n+3)/6] - 1 + Sn2 + Sn+22 + ..............
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23 Dec 2007 19:36:20 IST
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edited
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