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
\(A=1+\frac{1}{2}\cdot\left(1+2\right)+\frac{1}{3}\cdot\left(1+2+3\right)+...+\frac{1}{16}\cdot\left(1+2+3+...+16\right).\)
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
Tìm GTNN của 1+1/(1+2)+1/(1+2+3)+1/(1{2+3+4)+...+1/(1+2+3+...+n)