兩條數學題 急急急!!!!!! 請幫忙

1. 10000-9999+9998-9997+.....+4-3+2-1
= (10000-9999)+(9998-9997)+......(4-3)+(2-1)
= 1+1+.....+1+1
= 1 * 5000
= 5000

2. 10-9+8-7+6-5+4-3+2-1
= (10-9)+(8-7)+(6-5)+(4-3)+(2-1)
= 1+1+1+1+1
= 5

1. 10000-9999+9998-9997+.....+4-3+2-1
= (10000-9999)+(9998-9997)+......(4-3)+(2-1)
= 1+1+.....+1+1
= 1 * 5000
= 5000

2. 10-9+8-7+6-5+4-3+2-1
= (10-9)+(8-7)+(6-5)+(4-3)+(2-1)
= 1+1+1+1+1
= 5

1. 10000-9999+9998-9997+......+4-3+2-1
=(10000-9999)+(9998-9997)+......+(4-3)+(2-1)
=1+1+.....+1+1
=5000

2. 10-9+8-7+6-5+4-3+2-1
=(10-9)+(8-7)+(6-5)+(4-3)+(2-1)
=1+1+1+1+1
=5

In fact, we only need to use brackets to separate them into different groups and we can find out the answers of the two questions above easily.

*Be careful that a-b-c is not equal to a-(b-c).

第1題:5000
第2題:5

第一題係

((10000/2 * (10000+2)) / 2) -
(((9999+1)/2 * (9999+1)) / 2)

=5000

1. 5000
2. 5

第2題係5