Đáp án: $E=\dfrac{n+1}{2(3n+5)}$

Giải thích các bước giải:

$E=\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{(3n+2).(3n+5)}$

$3E=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{(3n+2).(3n+5)}$

Với phân số có dạng $\dfrac{3}{(3n+2).(3n+5)}$

$=\dfrac{(3n+5)-(3n+2)}{(3n+2).(3n+5)}$

$=\dfrac{3n+5}{(3n+2).(3n+5)}-\dfrac{3n+2}{(3n+2).(3n+5)}$

$=\dfrac{1}{3n+2}-\dfrac{1}{3n+5}$

Chứng minh tương tự với mọi phân số còn lại, có:

$3E=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n+2}-\dfrac{1}{3n+5}$

$3E=\dfrac{1}{2}-\dfrac{1}{3n+5}$

$3E=\dfrac{3n+5}{2(3n+5)}-\dfrac{2}{2(3n+5)}$

$3E=\dfrac{3n+5-2}{2(3n+5)}$

$3E=\dfrac{3n+3}{2(3n+5)}$

$3E=\dfrac{3(n+1)}{2(3n+5)}$

$E=\dfrac{3(n+1)}{2(3n+5)}:3$

$E=\dfrac{3(n+1)}{2(3n+5)}.\dfrac13$

$E=\dfrac{n+1}{2(3n+5)}$

Vậy $E=\dfrac{n+1}{2(3n+5)}$

Đáp án:

`E = 1/[2.5]+1/[5.8]+1/[8.11]+....+1/[(3n+2).(3n+5)]`

`E = 3/3 . (1/[2.5]+1/[5.8]+1/[8.11]+....+1/[(3n+2).(3n+5)])`

`E = 1/3 . (3/[2.5]+3/[5.8]+3/[8.11]+....+3/[(3n+2).(3n+5)])`

`E = 1/3 . (1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/(3n + 2) - 1/(3n + 5))`

`E = 1/3 . (1/2 - 1/(3n + 5))`

`E =1/3 .[3n+5−2]/[2.(3n+5)]`

`E =[3n+3]/[6.(3n+5)]`

`E =[3.(n + 1)]/[6.(3n+5)]`

`E = (n + 1)/(2.(3n + 5))`

$#dariana$