Question

[Nguyễn Mai Phương]

tính : 1+1\2*(1+2)+1\3*(1+2+3)+...+1\6*(1+2+3+...+16)

Answer 1

[Nguyễn Hưng Phát]

Cho n\(\in\)N*.CMR:

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

Ta có công thức:1+2+3+.....+n=\(\frac{n.\left(n+1\right)}{2}\)

Thật vậy:\(\frac{1}{n}.\left(1+2+.....+n\right)=\frac{n+1}{2}\)

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

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

\(\Rightarrow\frac{n+1}{2}=\frac{n+1}{2}\)(đúng)

Thay vào ta có:\(1+\frac{1}{2}.\left(1+2\right)+.......+\frac{1}{16}.\left(1+2+3+....+16\right)\)

=\(1+\frac{2+1}{2}+.....+\frac{16+1}{2}\)

=\(1+\frac{3}{2}+.......+\frac{17}{2}\)

=\(\frac{2+3+....+17}{2}\)

=\(\frac{152}{2}\)

=76

Answer 2

[Đinh Thùy Linh]

\(A=1+\frac{1}{2}\cdot\left(1+2\right)+\frac{1}{3}\cdot\left(1+2+3\right)+...+\frac{1}{16}\cdot\left(1+2+3+...+16\right).\)

Tổng của n số tự nhiên liên tiếp là: \(1+2+3+...+n=\frac{n\cdot\left(n+1\right)}{2}\). 

\(A=\frac{2}{2}+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+...+\frac{1}{16}\cdot\frac{16\cdot17}{2}.\)

\(=-\frac{1}{2}+\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)

\(=-\frac{1}{2}+\frac{1+2+3+4+...+17}{2}=-\frac{1}{2}+\frac{\frac{17\cdot18}{2}}{2}=-\frac{1}{2}+\frac{153}{2}=\frac{152}{2}=76\)

Đ/S A = 76