ধারার সমস্যা

For students of class 11-12 (age 16+)
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sm.joty
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ধারার সমস্যা

Unread post by sm.joty » Sun Oct 16, 2011 3:35 pm

একটা ধারা সংজ্ঞায়িত হয়েসে এভাবে,
\[a_{0} = 1
\]
\[a_{2n}=a_{2n+1}-a_{2n-1}
\]
\[a_{2n+1}=a_{n}
\]
তাহলে ,\[a_{2011} = ?
\]

কেউ কি আমাকে এই সমস্যাটা সমাধানে কোন help করতে পারবেন।

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Corei13
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Re: ধারার সমস্যা

Unread post by Corei13 » Sun Oct 16, 2011 6:53 pm

Hint :
$a_{4n+6} = a_{4n+7} - a_{4n+5}=a_{2n+3}-a_{2n+2}=a_{2n+1}=a_{n}$
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photon
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Re: ধারার সমস্যা

Unread post by photon » Sun Oct 16, 2011 7:05 pm

i found $-1$,with a big calculation.
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Akash
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Re: ধারার সমস্যা

Unread post by Akash » Sat Dec 10, 2011 12:49 pm

Is the answer 1?

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nafistiham
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Re: ধারার সমস্যা

Unread post by nafistiham » Sat Dec 10, 2011 1:57 pm

i think it to be $1$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Labib
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Re: ধারার সমস্যা

Unread post by Labib » Sat Dec 10, 2011 2:41 pm

The solution's $-1$ alright. But I'm trying to use Dj's hint still now...
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.


"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

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nafistiham
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Re: ধারার সমস্যা

Unread post by nafistiham » Sat Dec 10, 2011 3:00 pm

yap, it is $-1$ not $1$
but, could not understand what dj da meant
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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Corei13
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Re: ধারার সমস্যা

Unread post by Corei13 » Sat Dec 10, 2011 5:52 pm

$a_{4n+6} = a_{4n+7} - a_{4n+5}=a_{2n+3}-a_{2n+2}=a_{2n+1}=a_{n}$
And, given that, $a_{4n} = a_{4n+1}-a_{4n-1}$ and so $a_2 = a_3 - a_1 = a_1 - a_0 = a_0 - a_0 = 0$
$a_{2011} = a_{1005} = a_{502} = a_{124} = a_{125} - a_{123} = a_{62} - a_{61} = a_{14} - a_{30} = a_2 - a_6 = a_2 - a_0 = -a_0 = -1$
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