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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Summation-cool prob
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[Post New]12 May 2008 10:10:17 IST
    Subject: Summation-cool prob
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manja (0)

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can anybody do this prob


 


s1=1*2*3=6


s2=1*2*3+1*3*4+2*3*4+1*2*4


sn=????????


 


 


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[Post New]13 May 2008 22:06:02 IST
    Subject: Re:Summation-cool prob
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allamraju (3410)

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GOOD QUESTION.Is the ans n2(n+1)2(n-1)(n-2)/48?If correct,I will definitely explain.Plz reply or nudge me.
MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC.
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[Post New]14 May 2008 00:49:34 IST
    Subject: Summation-cool prob
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RyuAmakusa (461)

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i dont think that is correct. if u sub n=1 or n=2 it becomes 0 but we know that it is not.
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[Post New]14 May 2008 10:14:32 IST
    Subject: Re:Summation-cool prob
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sboosy (3009)

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\mbox{The value of n is greater than or equal to 3} \\ \\ \mbox{Thus substituting n as 1 or 2 is invalid} \\ \\ \mbox{Consider} \ S_n-S_{n-1} \\ \mbox{It can be easily shown to be equal to} \ \frac{n}{2} \left(\left(\frac{n(n-1)}{2}\right)^2 - \frac{n(n-1)(2n-1)}{6}\right) \\ \\ = \frac{n^2(n-1)(n-2)(3n-1)}{24} = K \\ \\ \mbox{Obtained using the fact that sum taken 2 at a time is} \ \frac{(1+2+3+..)^2-(1^2+2^2+3^2+..)}{2} \\ \\ \mbox{In K if we sub n as 4,we get} \ S_4-S_3 \\ \\ n=5 \ \Rightarrow S_5 - S_4 \\ \\ \Rightarrow \sum( \mbox{4 to n})K = S_n - S_3 \\ \\ \Rightarrow S_n = \sum( \mbox{4 to n})K+S_3
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