Question

Tính nhanh A=1+\(\dfrac{1}{2}\)(1+2)+\(\dfrac{1}{3}\)(1+2+3)+\(\dfrac{1}{4}\)(1+2+3+4)+...+\(\dfrac{1}{16}\)(1+2+3+...+16)

Answer 1

\(=1+\dfrac{1}{2}.\dfrac{2.3}{2}+\dfrac{1}{3}.\dfrac{3.4}{2}+...+\dfrac{1}{16}.\dfrac{15.16}{2}\)

\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{16}{2}\)

\(=\dfrac{1}{2}+\left(2+3+4+...+16\right)\) Trong ngoặc có (16-2):1+1=15 (số hạng)

\(=\dfrac{1}{2}+\dfrac{\left(16+2\right).15}{2}\)

\(=\dfrac{1}{2}.9.15=\dfrac{135}{2}=67\dfrac{1}{2}\)

Answer 2

nhận xét: \(\dfrac{1}{n}\left(1+2+...+n\right)=\dfrac{n\left(n+1\right)}{2n}=\dfrac{n+1}{2}\)

=>A=\(\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{16}{2}=\dfrac{\left(16-2+1\right)\cdot\left(16+2\right)}{2.2}=\dfrac{135}{2}\)

Answer 3

\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\\ =1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\\ =1+\dfrac{1\cdot2\cdot3}{2\cdot2}+\dfrac{1\cdot3\cdot4}{3\cdot2}+...+\dfrac{1\cdot16\cdot17}{16\cdot2}\\ =\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}\\ =\dfrac{2+3+4+...+17}{2}\\ =\dfrac{152}{2}\\ =76\)