For discussing Olympiad Level Number Theory problems
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*Mahi*
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by *Mahi* » Thu Dec 29, 2011 9:17 pm
Identity:
\[\sum^n_{i=0} f_i = f_{n+2}-1\]
And if I'm correct, the third number should be $21$.
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Tahmid Hasan
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by Tahmid Hasan » Fri Dec 30, 2011 11:31 am
my calculation
$n$th term $2(n^2+1)+3$
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Phlembac Adib Hasan
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by Phlembac Adib Hasan » Fri Dec 30, 2011 12:21 pm
What are you talking about?.I don't understand.Do you want to find $\sum_{k=1}^{n}2(k^2+1)+3$?
$\sum_{k=1}^{n}2(k^2+1)+3$
$\Rightarrow (2\sum_{k=1}^{n}k^2)+5n$
$\Rightarrow \frac {1} {3} [2n^3+3n^2+n]+5n $
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Tahmid Hasan
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by Tahmid Hasan » Fri Dec 30, 2011 3:18 pm
Phlembac Adib Hasan wrote:What are you talking about?.I don't understand.Do you want to find $\sum_{k=1}^{n}2(k^2+1)+3$?
i just showed the generalized form of every term.now it is quite easy to find the sum.(just like you did
![Mad :x](./images/smilies/icon_mad.gif)
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*Mahi*
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by *Mahi* » Fri Dec 30, 2011 9:42 pm
Tahmid Hasan wrote:my calculation
$n$th term $2(n^2+1)+3$
How?
Last edited by
*Mahi* on Sat Dec 31, 2011 11:52 am, edited 1 time in total.
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nafistiham
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by nafistiham » Fri Dec 30, 2011 11:47 pm
*Mahi* wrote:Tahmid Hasan wrote:my calculation
$n$th term $2(n^2+1)+3$
How in earth?
well i have found that we get the number quite rightly.what's the problem here ?
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*Mahi*
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by *Mahi* » Sat Dec 31, 2011 11:51 am
The difference are $4n+2$, so it can be easily derived. I meant that.
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