Tính tổng sau:
a. 1/1.2+1/2.3+1/3.4+…+1/a.(a+1)
b. 2/1.3+2/3.5+2/5.7+…+2/(2n+1).(2n+3)
c. 1/3+1/15+1/35+1/63+1/99
Tính tổng sau:
a. 1/1.2+1/2.3+1/3.4+…+1/a.(a+1)
b. 2/1.3+2/3.5+2/5.7+…+2/(2n+1).(2n+3)
c. 1/3+1/15+1/35+1/63+1/99
a) `1/1.2 + 1/2.3 + 1/3.4 +…+ 1/(a.(a+1))`
`=1 – 1/2 + 1/2 – 1/3 + 1/3 -1/4+…+ 1/a – 1/(a+1)`
`= 1- 1/(a+1)`
b) `2/1.3 + 2/3.5 + 2/5.7 +…+ 2/((2n+1)(2n+3))`
`=1/1 – 1/3 + 1/3 – 1/5+ 1/5- 1/7 +…+ 1/(2n+1)- 1/(2n+3)`
`=1 – 1/(2n+3)`
c) Đặt `A=1/3 + 1/15 + 1/35 + 1/63 + 1/99`
`A= 1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11`
`2A = 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11`
`2A= 1/1 – 1/3 + 1/3 -1/5 + 1/5- 1/7 + 1/7- 1/9 + 1/9 – 1/11`
`2A= 1 – 1/11`
`2A= 10/11`
`A= 10/11 :2`
`A= 10/11 . 1/2`
`A= 5/11`
Vậy `A= 5/11`
`a)` $\dfrac{1}{1.2}$ + $\dfrac{1}{2.3}$ + $\dfrac{1}{3.4}$ + …. + $\dfrac{1}{a(a+1)}$
` = 1/1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + … + 1/a – 1/(a+1)`
` = 1/1 + ( – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + … + 1/a ) – 1/(a+1) `
` = 1/1 + 0 – 1/(a+1) `
` =1/1 – 1/(a +1 ) = (a+1)/(a+1) – 1/(a+1) = a/(a+1)`
`b)` $\dfrac{2}{1.3}$ + $\dfrac{2}{3.5}$ + $\dfrac{2}{5.7}$ + … + $\dfrac{2}{(2n+1 )(2n+3)}$
`= 1/1 – 1/3 + 1/3 – 1/5 + 1/5 – 1/7 + …. + 1/(2n+1) – 1/(2n+3)`
`= 1/1 + 0 – 1/(2n+3)`
`= 1/1 – 1/(2n+3) = (2n+3)/(2n+3) – 1/(2n+3) = (2n+2)/(2n+3) `
`c) 1/3 + 1/15 + 1/35 + 1/63 + 1/99`
`= 1/(1.3) + 1/(3.5) + 1/(5.7) + 1/(7.9) + 1/(9.11)`
Đặt `B` = ` 1/(1.3) + 1/(3.5) + 1/(5.7) + 1/(7.9) + 1/(9.11)`
`2B = 2/(1.3) + 2/(3.5) + 2/(5.7) + 2/(7.9) + 2/(9.11) `
`2B = 1/1 – 1/3 + 1/3 – 1/5 + …. + 1/9 – 1/11 `
`2B = 1/1 – 1/11 = 10/11`
`→ B = 10/11 : 2 = 10/11 . 1/2 = 5/11`