ধারার সমস্যা
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।
(এটা Holly Cross Collage Olympiad এর সমস্যা। এটা officially upload করা হয়েছে কিনা আমার জানা নাই। )
\[a_{0} = 1
\]
\[a_{2n}=a_{2n+1}-a_{2n-1}
\]
\[a_{2n+1}=a_{n}
\]
তাহলে ,\[a_{2011} = ?
\]
কেউ কি আমাকে এই সমস্যাটা সমাধানে কোন help করতে পারবেন।
(এটা Holly Cross Collage Olympiad এর সমস্যা। এটা officially upload করা হয়েছে কিনা আমার জানা নাই। )
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তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
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বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
Re: ধারার সমস্যা
i found $-1$,with a big calculation.
Try not to become a man of success but rather to become a man of value.-Albert Einstein
- nafistiham
- Posts:829
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Re: ধারার সমস্যা
i think it to be $1$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: ধারার সমস্যা
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
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
- nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
- Location:24.758613,90.400161
- Contact:
Re: ধারার সমস্যা
yap, it is $-1$ not $1$
but, could not understand what dj da meant
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.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: ধারার সমস্যা
$a_{4n+6} = a_{4n+7} - a_{4n+5}=a_{2n+3}-a_{2n+2}=a_{2n+1}=a_{n}$
And, given that, $a_{4n} = a_{4n+1}-a_{4n-1}$ and so $a_2 = a_3 - a_1 = a_1 - a_0 = a_0 - a_0 = 0$
$a_{2011} = a_{1005} = a_{502} = a_{124} = a_{125} - a_{123} = a_{62} - a_{61} = a_{14} - a_{30} = a_2 - a_6 = a_2 - a_0 = -a_0 = -1$
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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