Đáp án:
`M = 4080399/2020`
Giải thích các bước giải:
Đặt `A^2 = 1+ 1/a^2 + 1/(a+1)^2= (a^2(a+1)^2+(a+1)^2 + a^2)/(a^2(a+1)^2)`
`= (a^2 (a^2 + 2a + 1 +1)+(a+1)^2)/(a^2(a+1)^2)`
`= (a^4 + 2a^2(a+1)+(a+1)^2)/(a^2(a+1)^2)`
`= (a^2 + a+1)^2/(a^2(a+1)^2)`
`= [(a^2+a+1)/(a(a+1))]^2`
Vì `a>0` nên `A>0` suy ra `A= (a^2 + a+1)/(a(a+1))`
Do đó `A = sqrt(1+1/a^2 + 1/(a+1)^2)`
`= (a^2+a+1)/(a(a+1))`
`= 1+1/(a(1+1))`
`= 1+1/a - 1/(a+1)`
Suy ra `M = (1+1/1-1/2) + (1+1/2-1/3)+...+ (1+1/2019-1/2020)`
`= 2019+(1/1-1/2+1/2-1/3+...+1/2019-1/2020)`
`= 2019 + (1-1/2020)`
`= 2020 - 1/2020`
`= (2020^2-1)/2020`
`= 4080399/2020`
