Trình bày lời giải:
Bài `24.`
`a)`
`A=1-3+5-7+9-11+...+97-99`
`A=\underbrace{(1-3)+(5-7)+(9-11)+...+(97-99)}_{\text{có [(99-1):2+1]:2=25 cặp}}`
`A=\underbrace{(-2)+(-2)+(-2)+...+(-2)}_{\text{có 25 số -2}}`
`A=-2.25`
`A=-50`
Vậy `A=-50`
`b)`
`B=-1-2-3-4-...-100`
`B=-(` `\underbrace{1+2+3+4+...+100}_{\text{Có (100-1):1+1=100 số}}` `)`
`B=-[(100+1)xx100:2]`
`B=-[101xx100:2]`
`B=-[10100:2]`
`B=-5050`
Vậy `B=-5050`
`c)`
`C=1-2+3-4+5-6+...+99-100`
`C=\underbrace{(1-2)+(3-4)+(5-6)+...+(99-100)}_{\text{Có [(100-1):1+1]:2=50 cặp}}`
`C=\underbrace{-1+(-1)+(-1)+...+(-1)}_{\text{Có 50 số -1}}`
`C=-1.50`
`C=-50`
Vậy `C=-50`
`d)`
`D=1-2-3+4+5-6-7+8+9-...-94-95`
`D=\underbrace{(1-2)+(-3+4)+(5-6)+...+(93-94)}_{\text{Có [(94-1):1+1]:2=47 cặp}}-95`
`D=-1+1-1+...-1-95`
`D=(-1+1)+(-1+1)+...+(-1+1)-1-95`
`D=-96`
Vậy `D=-96`
