MATHMANIA 2004!!!

Discus: SAT/ACT Tests and Test Preparation: July 2004 Archive: MATHMANIA 2004!!!
 By Loop123 (Loop123) on Tuesday, July 13, 2004 - 11:25 pm: Edit

You're a contestant on MATHMANIA 2004!!!

Let a_n mean "a with the subscript of n".

Given the sequence 1, 2, 4, 7, 11, 16, 22....

Create a general equation for finding the nth term of the sequence.

 By Geniusash (Geniusash) on Tuesday, July 13, 2004 - 11:28 pm: Edit

a_n=a_(n-1)+(n-1)

 By Loop123 (Loop123) on Tuesday, July 13, 2004 - 11:33 pm: Edit

*Double Post*

 By Loop123 (Loop123) on Tuesday, July 13, 2004 - 11:33 pm: Edit

That is a decent answer --- but there is a better one. Try to find an answer where the nth term can be found without the need to find a_(n-1).

 By Al0 (Al0) on Tuesday, July 13, 2004 - 11:49 pm: Edit

a_n=(n^2)/2+n/2+1

which gives the series: 1,2,4,7,11,16,22,29,37,46 . . .

 By Loop123 (Loop123) on Wednesday, July 14, 2004 - 03:14 pm: Edit

Any other takers for question #1 on MATHMANIA 2004!?

 By Benzinspeicher (Benzinspeicher) on Wednesday, July 14, 2004 - 03:24 pm: Edit

(n^2-n+2)/2

 By Benzinspeicher (Benzinspeicher) on Wednesday, July 14, 2004 - 05:02 pm: Edit

next?

 By Al0 (Al0) on Wednesday, July 14, 2004 - 05:32 pm: Edit

Benzinspeicher, your generator has an OBOB (because all you did was subtract n from mine and re-arrange) and yields 1,1,2,4,7,11,16. . . Remember that sequences begin with a0 (a-nought)

 By Jenesaispas (Jenesaispas) on Wednesday, July 14, 2004 - 05:40 pm: Edit

Um, here's a function that isn't the most accurate one. But it's okay, we can pretend it is. And we can round to the nearest... one, if that's okay.
But it only works for the first 899,999 terms.

Er, a_n=(-.00000000000005126953125011)(n^4) + (.00000000000092285156250198)(n^3) + (.49999999999443)(n^2) - (.4999999999872)(n) + .999999999991

 By Loop123 (Loop123) on Wednesday, July 14, 2004 - 06:12 pm: Edit

The next problem also deals with finding the general equation for the nth term of a sequence.

Given the sequence

1, 1/2, 1/6, 1/24...

Derive a general equation for finding the nth term of the sequence.

 By Firebird12637 (Firebird12637) on Wednesday, July 14, 2004 - 06:19 pm: Edit

partial fractions:

nth term, where n > 1 = 1/(n-1) - 1/n

 By Loop123 (Loop123) on Wednesday, July 14, 2004 - 06:21 pm: Edit

Try to find an equation where n can equal one and it still holds true. I'll give you a hint!!!!!!!!

 By Al0 (Al0) on Wednesday, July 14, 2004 - 06:26 pm: Edit

a_n=1/[(x+1)!] gives 1,1/2,1/6,1/24,1/120 . . .

 By Benzinspeicher (Benzinspeicher) on Wednesday, July 14, 2004 - 09:05 pm: Edit

Al0, i didn't even see yours as i solved it, yours is probably the same as mine

 By Benzinspeicher (Benzinspeicher) on Wednesday, July 14, 2004 - 09:09 pm: Edit

would that be 1/n! ?

 By Al0 (Al0) on Wednesday, July 14, 2004 - 11:57 pm: Edit

a_n=1/[(n+1)!] but you have to have the n+1, not simply n (or you get an OBOB) or my x+1 (which would be constant).

 By Benzinspeicher (Benzinspeicher) on Friday, July 16, 2004 - 08:20 pm: Edit

any more problems?

 By Loop123 (Loop123) on Friday, July 16, 2004 - 08:33 pm: Edit

2 cells of a 7 x 7 board are painted black and the rest white. How many different boards can be produced? (boards which can be rotated into each other do not count as different).

 By Al0 (Al0) on Saturday, July 17, 2004 - 10:24 pm: Edit

I have two solutions:

1. The Hard Way (more fun, though):
Step 1: Draw 7x7 Board
Step 2: Label cell(4,4) 0.
Step 3: Label the un-labeled cell nearest the origin with the next highest integer and the letter a.
Step 4: Label the 90 degree (the board is rectangular so only these need be considered) rotations of the cell in Step 3 about 0 with the same integer and the next letter.
Step 5: Repeat Steps 3 and 4 until all the cells are labeled.
*|_1_|_2_|_3_|_4_|_5_|_6_|_7_|
1|12d|11d|9d_|7a_|8a_|10a|12a|
2|10d|6d_|5d_|3a_|4a_|6a_|11a|
3|8d_|4d_|2d_|1a_|2a_|5a_|9a_|
4|7d_|3d_|1d_|_0_|1b_|3b_|7b_|
5|9c_|5c_|2c_|1c_|2b_|4b_|8b_|
6|11c|6c_|4c_|3c_|5b_|6b_|10b|
7|12c|10c|8c_|7c_|9b_|11b|12b|

Looking at the table one can soon determine the following:
1. 0 forms only one unique pair with each number.
2. Homogeneous pairs only form two unique pairs (ab and ac).
3. Heterogeneous pairs form 4 unique pairs (aa,ab,ac,ad).
So the total number of pairs is 12 (from the 12 pairs involving 0) + 2*12 (from the 12 homogeneous pairs: 11,22,33,...,1212) + 4*(11+10+9+...+1) (from the heterogeneous pairs: 1 forms 11 such pairs [1 with each other number], 2 forms 10 such pairs [1 with each number except 1 because they were counted in the last set], etc.) for a grand total of:
12+2*12+4*(11+10+9+8+7+6+5+4+3+2+1)=300

2. The Easy Way
There are 49 choices for the first cell, there are then 48 choices for the second cell -- giving 49*48 order-dependent cell selections. Because the order doesn't matter, we have that number, so there are 49*48/2=1,176 pairs without accounting for rotational repeats. From the diagram above we saw that there are 12 rotational loops with 4 cells each and therefore 12*4/2=24 pairs which are double-counted when not accounting for rotation, with the other 1,176-24=1,152 being quadruple counted. Accounting for these duplicates, we get 1152/4(the quadruple counted)+24/2(the double counted)=288+12=300.