Math Challenge!!!!?

Question

Hi!!! Could you help solve one of these math challenge??? With solutions please!!! (there's like an easier way to get these than just calculate all of it):

1. (2-2/2)*(2-2/3)*(2-3/4)*...............(2-2004/2005)
2. 1+2-3+4+5-6+7+8-9+...........+601+602-603+604+605-606
3. 100-98+96-94+92-90+..........+8-6+4-2
4. (2+4+6+.....+100) - (1+3+5+.......+99)

Answer 1

1) With the later correction, starting with (2 - 1/2) * ... * (2 - 2004/2005), you're multiplying 3/2 * 4/3 (cancels out the 3 and leaves 4/2), then by 5/4 (cancels out the 4 and leaves 5/2), etc. The last term (2 - 2004/2005) is 2006/2005, which will cancel out 2005 in the numerator and replace it with 2006. The answer is 2006/2 which is the same as 1,003.



If the final denominator in the series is "n", the answer is (n+1)/2.



2) The terms in groups of 3 add up to 0 (1+2-3) + 3 (4+5-6) + 6 (7+8-9) + 9 + 12 + ... and so on. Yahoo answers truncated your stopping point (you have to put spaces every 30 or so characters or it gets truncated) so I'm not sure where you wanted the series to end.



If you wanted it to end at 99 (just a guess), since your other series involve numbers up to about 100, then the final terms would be 97+98-99, which as a group of 3 add up to 96. The sum of 3 + 6 + 9 + 12 + ... + 96 is the average of the first and last term (3 + 96)/2 = 49.5, times the number of terms (32), which is 1,584. (If you want to stop at 100, the answer is 1,684, but 100 isn't part of a group of 3 terms.)



3) 100-98 = 2, 96-94 = 2, etc., so this one reduces to 2 + 2 + 2 + ... Assuming you stop at + 4 - 2 (+ 0), there are 50 even numbers between 2 and 100 inclusive. That gives us 25 PAIRS of terms down to 4-2, so assuming you stop there and don't get into negative numbers, each pair adds two to the sum, so the answer is 25*2, which is 50.



4) Each of the 50 numbers in the first set is one more than the corresponding number in the second set. So this answer is 50 as well.

Answer 2

1.

2/2 = 1

2-2/2 = 2-1 = 1



since multiplying by 1 does nothing, we can leave that first term (2-2/2) out of the problem and get the same answer.



for the rest of the series:

there is a pattern in the numbers being subtracted

2-2/3 = 6/3-2/3 = 4/3

2-3/4 = 8/4-3/4 = 5/4

2-4/5 = 10/5-4/5 = 6/5

...

2-2004/2005 = 4010/2005 - 2004/2005 = 2006/2005



so the product is

(4/3)*(5/4)*...*(2006/2005)

= (4*5*...*2006)/(3*4*...*2005)



so basically, on the top you get 2006!/6 and on the bottom you get 2005!/2



but 2006! = 2005!*2006



(2006!/6) /(2005!/2) = (2005!*2006/2) / (2005!/6)

= (2006/2) / (1/6)

= 6 * (2006/2)

= 3 * 2006

= 6018



2.

(1+2-3) = 0 = 3*0

+

(4+5-6) = 3 = 3*1

+

(7+8-9) = 6 = 3*2

+

(10+11-12) = 9 = 3*3



They are multiples of 3 for every 3 numbers.



So since it goes up to 606, 606/3 = 101 groups of 3, but since the first group of three is = 0 = (1+2-3), we'll just say 100 groups of 3.



so you get

(3*1)+(3*2)+...+(3*99)+(3*100)

3(1+2+3+...+99+100)

the sum from 1 to 100 = 100(1+100)/2 = 5050

3*(5050) = 15150



3. (100-98) + (96-94) + ... + (4-2)

each of these groups in parentheses = 2

there are 50 even numbers between 2 and 100, including 2 and 100. And each of these groups uses 2 of those numbers so there are 50/2 = 25 groups.

So you're adding 2 to itself 25 times, the definition of multiplication.

2*25 = 50



4.

(2+4+6+...+100) - (1+3+5+...+99)



I can rewrite this



(2-1)+(4-3) + (6-5)+...+(100-99).

each of these pairs in parentheses = 1

since we go through 100 numbers in pairs of 2, there are 100/2 or 50 groups in parentheses. So basically, you add 1 to itself 50 times.



So the answer is 50.

Answer 3

Assume the first three series are infinite series.

1. It seems to me that the first chain should be (2-1/2)

(2-1/2)*(2-2/3)*(2-3/4)*.........

= (3/2)(4/3)(5/4)(6/5)...

= n/2 ->∞, as n->∞, where n stands for the nth factor.



2.

1+2-3+4+5-6+7+8-9+...........+...

= (1+2-3)+(4+5-6)+(7+8-9)+...->+∞, as n->∞ because each succeeding term becomes larger and larger.



3. (100-98)+(96-94)+(92-90)+..........+...

=2+2+2+...->+∞, as n->∞



4.

(2+4+6+.....+100) - (1+3+5+.......+99)

= (2-1)+(4-3)+...+(100-99)

= 50, since you have 50 pairs.