Đáp án:

$S_4=\dfrac{297}{1000};S_5= \dfrac{99}{2000}.$

Giải thích các bước giải:

Ta có: $\dfrac{2}{n(n+2)}=\dfrac{n+2-n}{n(n+2)}=\dfrac{n+2}{n(n+2)}-\dfrac{n}{n(n+2)}=\dfrac{1}{n}-\dfrac{1}{n+2}$

$S_4=\dfrac{6}{10.12}+\dfrac{6}{12.14}+\dots+\dfrac{6}{998.1000}\\ =3\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+\dots+\dfrac{2}{998.1000}\right)\\ =3\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+\dots+\dfrac{1}{998}-\dfrac{1}{1000}\right)\\ =3\left(\dfrac{1}{10}-\dfrac{1}{1000}\right)\\ =3\left(\dfrac{100}{1000}-\dfrac{1}{1000}\right)\\ =3.\dfrac{99}{1000}\\ =\dfrac{297}{1000}\\ S_5=\dfrac{1}{10.12}+\dfrac{1}{12.14}+\dots+\dfrac{1}{998.1000}\\ =\dfrac{1}{2} \left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+\dots+\dfrac{2}{998.1000}\right)\\ =\dfrac{1}{2} \left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+\dots+\dfrac{1}{998}-\dfrac{1}{1000}\right)\\ =\dfrac{1}{2} \left(\dfrac{1}{10}-\dfrac{1}{1000}\right)\\ =\dfrac{1}{2} \left(\dfrac{100}{1000}-\dfrac{1}{1000}\right)\\ =\dfrac{1}{2} \dfrac{99}{1000}\\ = \dfrac{99}{2000}.$

Lời giải:

Sửa đề câu 1:

`S_4 = 6/(10 * 12) + 6/(12 * 14) +...+ 6/(998 * 1000)`

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`S_4 = 6/(10 * 12) + 6/(12 * 14) +...+ 6/(998 * 1000)`

      `= 6/2(2/(10 * 12) + 2/(12 * 14) +...+ 2/(998 * 1000))`

      `= 3(1/10 - 1/12 + 1/12 - 1/14 +...+ 1/998 - 1/1000)`

      `= 3(1/10 - 1/1000)`

      `= 3 * 99/1000`

      `= 297/1000`

`S_5 = 1/(10 * 12) + 1/(12 * 14) +...+ 1/(998 * 1000)`

      `= 1/2(2/(10 * 12) + 2/(12 * 14) +...+ 2/(998 * 1000))`

      `= 1/2(1/10 - 1/12 + 1/12 - 1/14 +...+ 1/998 - 1/1000)`

      `= 1/2(1/10 - 1/1000)`

      `= 1/2 * 99/1000`

      `= 99/2000`