Question

Tính P biết :

\(p=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+.....+\frac{1}{16}\left(1+2+...+16\right)\)

Answer 1

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

Do đó

P = \(1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+16}{16}\)

\(P=1+\frac{2.3}{2.2}+\frac{3.4}{2.3}+\frac{4.5}{2.4}+...+\frac{16.17}{2.16}\)

\(P=1+\frac{1}{2}\left(3+4+5+...+17\right)\)

\(P=1+\frac{1}{2}.\frac{\left(17-3+1\right)\left(3+17\right)}{2}=76\)

Answer 2

Xét thừa số tổng quát:

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

Thay vào bài toán:

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

\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}=\frac{2+3+...+17}{2}=76\)