Question

Chứng tỏ rằng: \(\dfrac{1}{a}\) - \(\dfrac{1}{a+1}\) = \(\dfrac{1}{a.\left(a+1\right)}\)

Tính: \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{99.100}\)

Answer 1

Ta có: \(\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{a+1}{a\left(a+1\right)}-\dfrac{a}{a\left(a+1\right)}\)

\(=\dfrac{a+1-a}{a\left(a+1\right)}\)

\(=\dfrac{1}{a\left(a+1\right)}\) (đpcm)

Ta được:

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+...-\dfrac{1}{100}\) \(=1-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)