Question

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{99}{100}\)

Answer 1

\(\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{99}{100}\)

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

\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{1}{n+1}=\dfrac{1}{100}\)

=> n+1=100

n=99

Answer 2

1/2 + 1/6 + 1/12 + ... + 1/(n(n+1)) = 99/100`

`=> 1/(1.2) + 1/(2.3) + 1/(3.4) + ... + 1/(n.(n+1)) = 99/100`

`=> 1/1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + ... + 1/n - 1/(n+1) = 99/100`

`=> 1 - 1/(n+1) = 99/100`

`=>      1/(n + 1) = 1 - 99/100`

`=>    1/(n + 1 ) = 1/100`

`=>       n  +1   = 100`

`=>      n          =  99`