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Alex
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Category: Homework
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Experience:  Tutoring college students since 1996. MS in CompSci 1996, MS in Math 2001.
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1 1 2 4 3 10 4 20 5 35 6 56 7 84 8 120 9 165

Customer Question

1=1
2=4
3=10
4=20
5=35
6=56
7=84
8=120
9=165
10=220
what is an equation that will work for all of them
Submitted: 10 years ago.
Category: Homework
Expert:  Lauren N replied 10 years ago.
Hi there!
Lets say our first column is N, and our second column is Y, so

Yn= N^2 + Y(n-2)
(this means that the y that corresponds to the n is equal to N squared plus the y that was 2 places ago; for example, if we want to look at N=5, then it will be Y5= 5^2 plus Y (5-2) = 5^2 + Y3=5^2 plus the number that is in the Y3 position which is 10; so its 5^2 plus 10)

For n=2, it will be 2^2 plus the number that was 2 places ago, there was none, so its 0, which is 4 plus 0, which equals 4. 3^2 plus 1 (it was 2 places ago) which is 10, and so on! for n=4, its 4^2 plus 4(which was 2 places ago) which is 16 plus 4 which is 20, and this works for all of them so this works out! You just go through, square the N, look 2 places back in the y column, add that to N squared and you get your new Y. (10^2 plus 2 places ago in the y position (120) is 100 plus 120 which is 220. I hope this makes sense!

Hope this helped out! thanks for posting! if i helped, please accept! :)
Expert:  Alex replied 10 years ago.

Hello and welcome to JA!


The answer is n3/6 + n2/2 + n/3. Here's my solution:



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Reference:
http://eqworld.ipmnet.ru/en/solutions/fe/fe1202.pdf


Let me know if you need more help with this question.



Expert:  Alex replied 10 years ago.
THIS ANSWER IS LOCKED!

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