Đáp án:
Giải thích các bước giải:
a/ $D=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+......+\dfrac{1}{19.20}$
⇔ $D=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{19}-\dfrac{1}{20}$
⇔ $D=\dfrac{1}{2}-\dfrac{1}{20}$
⇔ $D=\dfrac{10}{20}-\dfrac{1}{20}$
⇔ $D=\dfrac{9}{20}$
b/ $E=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}$
⇔ $E=\dfrac{1}{99}-(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{97.98}+\dfrac{1}{98.99})$
⇔ $E=\dfrac{1}{99}-(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99})$
⇔ $E=\dfrac{1}{99}-(1-\dfrac{1}{99})$
⇔ $E=\dfrac{1}{99}-1+\dfrac{1}{99}$
⇔ $E=\dfrac{2}{99}-1$
⇔ $E=-\dfrac{97}{99}$
$\text{Quy tắc cho bài trên:}$
$\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$
$\text{Quy tắc chung:}$
$\dfrac{1}{k.(k+a)}=\dfrac{1}{a}.(\dfrac{1}{k}-\dfrac{1}{k+a})$
Chúc bạn học tốt !!!
