Câu hỏi của Phan Nhật Hiếu

Question

1 + 12</sp...

Nguyễn Lê Phước Thịnh · Accepted Answer

$1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+20}$

$=1+\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{20\cdot\dfrac{21}{2}}$

$=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{20\cdot21}$

$=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{20\cdot21}\right)$

$=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)$

$=2\left(1-\dfrac{1}{21}\right)=2\cdot\dfrac{20}{21}=\dfrac{40}{21}$

Câu trả lời:

$1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+20}$

$=1+\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{20\cdot\dfrac{21}{2}}$

$=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{20\cdot21}$

$=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{20\cdot21}\right)$

$=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)$

$=2\left(1-\dfrac{1}{21}\right)=2\cdot\dfrac{20}{21}=\dfrac{40}{21}$