[ Arithmetic operations. Numerical identities ] [ Numerical inequalities. Comparing numbers. ]

Which of the numbers is greater: $1-2+3-4+5-\ldots+99-100$ or $1+2-3+4-5+6-\ldots-99+100$ ?
The first method. The first number is $(1-2)+(3-4)+(5-6)+\ldots+(99-100)=-50$, since the difference of the numbers in each parenthesis is $(-1)$,

and there are 50 such parentheses in total. Similarly, the second number is $(1+2)+(-3+4)+\ldots+(-99+100)=3+49=52$.

Therefore, the second number is greater than the first.

The second method. By combining the terms in pairs as above, we get that the first number is negative (the sum of negative numbers). The sum of the two given numbers is $(1-2+3-4+\ldots+99-100)+(1+2-3+4-\ldots$ $-99+100)=1+1=2$, (all other terms cancel each other out). Since the sum of the two numbers is positive, and the first number is negative, the second number is positive and, therefore, greater than the first.

## Answer

The second number is greater