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. 1+2-3+4+5-6+7+8-9+...........+...
3. 100-98+96-94+92-90+..........+...
4. (2+4+6+.....+100) - (1+3+5+.......+99)

Sorry for 1 and 2. I don't know why but i put the continuation of it..... wll, for number 1. ......... (2-2004/2005)

Answer 1

The only ones you can solve are 3 and 4 because 1 and 2 have no limits
1) Assuming that the sequence goes to (2-99/100), This can be easily solved. Simplifying each number in the series results in 1*4/3*5/4*6/5.... 101/100. The numerators cancel out with the denominators for each number in the series. Thus
it can be read as 1*5/3*6/5...-> 1*6/3* 7/6... all the way to 101/100. The 1/3 never gets canceled out,Thus the answer is 101/3. Going on to infinity however, the sum would be 1/3. The 1/3 would never get cancelled out and the other numerators/denominators would.
2) Assuming that the sequence goes to a 100, you can use the summation formula n(n+1)/2 Where n is from the sequence 1+2+3...n. Thus the answer is 100(101)/2 or 5050
If the sequence goes to some other number, it can still be solved with the formula.
3) After adding 100 and subtracting 98, you actually are adding 2. This is the same for other numbers. So the expression could read 2+2+2+2+2... Counting the even numbers from 0 to a 100 you get 50. Thus you are adding 2 50 times. So the answer is 100

4) You can use the addition formula. For the first part, the part is really 2(1+2+3+4+5...50). From the formula, n(n+1)/2 where n is the last number. Thus the answer to the first part is 2(50*51/2) or 2550. The second part is a little tricky. The second series can be seperated out to (1+(1+2)+(1+4)....(1+98)) The 1's all add up to 50, as there are 50 odd numbers from 0 to a 100. Thus the summation becomes 50+(2+4+6+...98) or 50+2(1+2+3...49). The sequence 2(1+2...49) is a 100 less than 2(1+2...50). Thus the second sequence answer comes out to 2450+50 or 2500. Thus the final answer is 2550-2500 or 50

Note: no 4 can also be done like no 3. By adding the 2 series you get -1+2-3+4....100. This is actually 1+1+1+1+1.... As 2-1 is 1, 4-3 is 1 etc... Thus you are adding one 50 times. The answer is 50

Answer 2

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.

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, it's 1684, 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 3

1.
(2-2/2)*(2-2/3)*(2-3/4)*.........
=2/2*4/3*5/4*6/5*7/6*........
=1*1/3*4*5/4*6/5*7/6*......
=1/3(all other cancels each other in numerator and denominator successively)

2.
1+2-3+4+5-6+7+8-9+...........+...
=1 -1 + 4 -1 + 7 -1 +.....
=Take all (-1)s together and other terms on another side
= (1 + 4 + 7 +10 +...n times) - (1+1+1....n times)
= (A.P. with first term 1 and common difference 3) - (n times 1)
= n/2(2+3*(n-1)) - n
= (n(2+ 3n-3) -2n)/2
=(3n^2 -3n)/2
=3n(n-1)/2

3.
100-98+96-94+92-90+..........+...
= 2 +2 +2 +2..........(two terms are giving each 2..therefore there are a total of 50 2s)
= 2+2+2+2...50 times
= 2*50
=100

4.
(2+4+6+.....+100) - (1+3+5+.......+99)
= Groups each terms from either side respectively
= (2-1) + (4-3)+(6-5)+(..)...+(100-99)
= 1 + 1 +1....+1 (here also two terms are giving each 1)
= 1 * 50 times
=50

Answer 4

2. 10 + 11
3. + 88 - 86
4. (2 + 4 + 6 + 94 + 100) - (1 + 3 + 5 + 94 + 99)

Answer 5

it's 0