Question

1+\(\frac{1}{2}\)*(1+2)+\(\frac{1}{3}\)*(1+2+3)+\(\frac{1}{4}\)*(1+2+3+4)+...........+\(\frac{1}{20}\)*(1+2+3.....+20)

Answer 1

Ta có: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\) áp dụng vào bài toán ta có

\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{4}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+...+20\right)\)

\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)

\(=\frac{1}{2}\left(2+3+4+...+20\right)=\frac{1}{2}.\frac{19.22}{2}=\frac{209}{2}\)

Answer 2

Ta có công thức :

1 + 2 + 3 + ... + n = \(\frac{n\left(n+1\right)}{2}\)

Áp dụng vào bài toán ta được :

\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)

\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{20}.\frac{20.21}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)

\(=\frac{2+3+4+...+21}{2}=\frac{\frac{21.22}{2}-1}{2}=115\)