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លំហាត់​នព្វន្ត៖ សិស្សពូកែរាជធានី ២០១០ (៤) ខែ​មីនា 15, 2010

Posted by វិចិត្រ in គណិតវិទ្យា.
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ចូរពិនិត្យស្វ៊ីតខាងក្រោម
u_1=1
u_2=3+5
u_3=7+9+11
u_4=13+15+17+19
….
ចូរបង្ហាញថា គ្រប់តួទាំងអស់នៃ {u_n } ជាចំនួនគូប។

ដំនោះ​ស្រាយ​របស់​ខ្ញុំ

មតិ»

1. ស៊ិម ទេពមុនី - ខែ​មីនា 24, 2010

Dear Bang Vichet,
I am glad to see your post. I wait for it for a long time…….
I have another way for solving this problem by first observing the terms of the sequences. Let check:
u_1 = 1 = 1;
u_2 = 3 + 5 = (2^2-1) + (2^2 +1) = 2^3;
u_3 = 7 + 9 +11 = (3^2-2) +3^2 + (3^2+2) = 3^3;
……………………………………………………………………
By this way, we can write kth term of the sequence, simply,
u_k = [k^2-(k-1)] + [k^2-(k-2)] +….+ [k^2+(k-2)] +[k^2 +(k-1)], which involves k terms. u_k is simplified to
u_k = K^2 + k^2 +….+ K^2 = k^3, where the other elements are vanish when calculating. Thus for any natural number k, u_k = k^3, which a cube, hence complete the proof.

2. ស៊ិម ទេពមុនី - ខែ​មីនា 24, 2010

Dear Bang Vichet,
I am glad to see your post. I wait for it for a long time…….
I have another way for solving this problem by first observing the patterns of the sequences. Let check:
u_1 = 1 = 1^3;
u_2 = 3 + 5 = (2^2-1) + (2^2 +1) = 2^3;
u_3 = 7 + 9 +11 = (3^2-2) +3^2 + (3^2+2) = 3^3;
……………………………………………………………………
By this way, we can write kth term of the sequence, simply,
u_k = [k^2-(k-1)] + [k^2-(k-2)] +….+ [k^2+(k-2)] +[k^2 +(k-1)], which involves k terms. u_k is simplified to
u_k = K^2 + k^2 +….+ K^2 = k^3, where the other elements vanish when calculating. Thus for any natural number k, u_k = k^3, which a cube, hence complete the proof.

ស៊ិម ទេពមុនី - ខែ​មីនា 24, 2010

There is a small mistake:
u_k = [k^2-(k-1)] + [k^2-(k-3)] +….+ [k^2+(k-3)] +[k^2 +(k-1)] is correct!!!

ស៊ិម ទេពមុនី - ខែ​មីនា 24, 2010

How can I remove the mistaken comments from my post?

វិចិត្រ - ខែ​មីនា 24, 2010

you can do it if you are member of this blog

ស៊ិម ទេពមុនី - ខែ​មីនា 24, 2010

Thank you very much, Bang Vichet. I attempted to find the application form for membership yesterday, however, it was impossible. Actually, I have emailed to the Group and Darareaksmey already, still I have not yet received any reply. I hope they will reply me. I want to be one of your members so that I can share my knowledge and some experience I have to next generation students. Working together is very efficient, I believe!

Mony
Math-Lec, ITC
EM-QEM.

3. តារារស្មី - ខែ​មីនា 24, 2010

Dear Mony, I am really sorry, but i really haven’t received any form from you, you may resend it! Anyway, when did u send?

4. តារារស្មី - ខែ​មីនា 24, 2010

Pls download the form and complete, ( find in a few previous posts )


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