\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + .....+\(\dfrac{1}{n.(n+1)}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\) +......+ \(\dfrac{1}{n}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{3}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{n+1}\) = \(\dfrac{1}{3}-\dfrac{3}{10}\)
\(\dfrac{1}{n+1}\) = \(\dfrac{1}{30}\)
n + 1 = 30
n = 30 - 1
n = 29
Kết luận n = 29 là giá trị thỏa mãn yêu cầu đề bài.