Question

chứng minh rằng:100-\(\left(1+\frac{1}{2}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)

Answer 1

\(100-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)=\left(1-1\right)+\left(1-\frac{1}{2}\right)+...+\left(1-\frac{1}{100}\right)\)

\(=0+\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)

\(\RightarrowĐPCM\)

 

Answer 2

Để chứng minh 100 - \(\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}\)

Thì ta cần chứng minh 100 = \(\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)+\left(\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}\right)\)

Biến đổi 

Vế phải = \(1+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{2}{3}\right)+....+\left(\frac{1}{100}+\frac{99}{100}\right)\)

            = 1 + 1 + 1 + ...1 (100 số 1) = 100 = Vế trái (đpcm)