Question

Tính nhanh

\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+...+\left(1+2+3+....+100\right)}{100.1+99.2+98.3+...+2.99+1.100}\)

dấu \(.\)là dấu x nhé

Answer 1

\(\frac{1+\left[1+2\right]+\left[1+2+3\right]+...+\left[1+2+3+...+100\right]}{100.1+99.2+98.3+...+2.99+1.100}=\frac{1.2:2+2.3:2+3.4:2+...+100.101:2}{100.1+99.2+98.3+...+2.99+1.100}\)

\(=\frac{\frac{1}{2}\left[1.2+2.3+3.4+...+100.101\right]}{100.1+99.2+98.3+...+2.99+1.100}=\frac{\frac{1}{2}\cdot\frac{1}{3}\left[1.2.3-0.1.2+2.3.4-1.2.3+...+100.101.102-99.100.101\right]}{1.100+2.100-1.2+3.100-2.3+...+100.100-99.100}\)

\(=\frac{\frac{1}{6}\cdot100.101.102}{100\left[1+2+3+...+100\right]-\left[1.2+2.3+...+99.100\right]}=\frac{171700}{100\cdot\frac{100.101}{2}-\frac{99.100\cdot101}{3}}\)

\(=\frac{171700}{505000-333300}=\frac{171700}{171700}=1\)

AI THẤY ĐÚNG NHỚ ỦNG HỘ NHÉ