Tìm `y` biết: `1/4+1/28+1/70+1/130+...+1/(y(y+3))=34/103` `1/3+1/6+1/10+1/15+...+1/((y(y+1))/2)=2009/2011`

Đáp án:

$a/$ `1/4 + 1/28 + 1/70 + 1/130 + .. + 1/(y (y + 3) ) = 34/103`

`⇔ 1/(1 . 4) + 1/(4 . 7) + 1/(7 . 10) + 1/(10 . 14) + ... + 1/(y (y + 3)) = 34/103`

`⇔ 1/3 [1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/14 + ... + 1/y - 1/(y + 3)] = 34/103`

`⇔ 1/3  [1  + ( - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/14 + ... + 1/y) - 1/(y +3)] = 34/103`

`⇔ 1/3 [1 -   1/(y + 3)] = 34/103`

`⇔ 1-  1/(y + 3) = 102/103`

`⇔ 1/(y + 3) = 1/103`

`⇔ 103 = y + 3`

`⇔ y = 100`

`text{Vậy y =100}`

$b/$ `1/3 + 1/6 + 1/10 + 1/15 + .. + 1/( (y (y + 1) )/2) = 2009/2011`

`⇔ 1/(2 . 3) + 1/(3 . 4) + 1/(4 . 5) + ... + 1/(y (y + 1) ) = 2009/4022`

`⇔ 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + .. + 1/y - 1/(y + 1) = 2009/4022`

`⇔ 1/2 + (- 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + .. + 1/y) - 1/(y + 1) = 2009/4022`

`⇔ 1/2 - 1/(y + 1) = 2009/4022`

`⇔ 1/(y + 1) = 1/2011`

`⇔ y + 1 = 2011`

`⇔ y = 2010`

`text{Vậy y = 2010}`