Bạn tham khảo câu trả lời tương tự ở đây nhé:
Câu hỏi của Nguyễn Hải - Toán lớp 7 - Học toán với OnlineMath
\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+...+\(\frac{2}{n\left(n+1\right)}\)=\(\frac{2017}{2019}\)
\(\frac{2}{6}\)+\(\frac{2}{12}\)+\(\frac{2}{20}\)+...+\(\frac{2}{n\left(n+1\right)}\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{n.\left(n+1\right)}\)\()\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2}\)_\(\frac{1}{3}\)+\(\frac{1}{3}\)_\(\frac{1}{4}\)+\(\frac{1}{4}\)_\(\frac{1}{5}\)+...+\(\frac{1}{n}\)_\(\frac{1}{n+1}\)\()\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2}\)_\(\frac{1}{n+1}\)\()\)=\(\frac{2017}{2019}\)
\(\frac{1}{2}\)_\(\frac{1}{n+1}\)=\(\frac{2017}{4038}\)
\(\frac{1}{n+1}\)=\(\frac{1}{2}\)_\(\frac{2017}{4038}\)
\(\frac{1}{n+1}\)=\(\frac{1}{2019}\)
\(\Rightarrow\)n+1=2019
\(\Rightarrow\)n=2018\(\in\)Z
Vậy n=2018
1/3 + 1/6 +1/10 +....+ 2/n(n+1) = 2017/2019
= 2/6 + 2/12 + 2/20 +...+ 2/n(n+1) = 2017/2019
= 2/2.3 + 2/3.4 + 2/4.5 +...+ 2/n.(n+1) = 2017/2019
= 2 .[ 1/2.3 + 1/3.4 + 1/4.5 +...+ 1/n.(n+1)] = 2017/2019
= 2 .(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/n - 1/n+1) = 2017/2019
= 2 .(1/2 - 1/n+1) = 2017/2019
= 1/2 - 1/n+1 = 2017/2019 : 2
= 1/2 - 1/n+1 = 2017/2019 . 1/2
= 1/2 - 1/n+1 = 2017/4038
= 1/n+1 = 1/2 - 2017/4038
= 1/n+1 = 2019/4038 - 2017/4038
= 1/n+1 = 1/2019
=> n+1 =2019
n = 2019 - 1
n = 2018
vậy n=2018
