1. Tìm số tự nhiên n sao cho :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{n.\left(n+1\right)}=\dfrac{2999}{3000}\)
2. Tính :
a ) \(S=2018.3+2018.4+2018.5+...+2018.2018\)
b ) \(\dfrac{1}{\sqrt{8}+\sqrt{10}}+\dfrac{1}{\sqrt{10}+\sqrt{12}}+\dfrac{1}{\sqrt{12}+\sqrt{14}}+...+\dfrac{1}{\sqrt{200}+\sqrt{202}}\)
c ) \(S=5.21^2+5.21^3+5.21^4+....+5.21^{2018}\)
d ) \(A=9+99+999+9999+...+9..9\)( 99 chữ số 9)
e ) 72+772+7772+...+77...72( 77 chữ số 7 )
2. Tính tổng :
a ) \(S=\dfrac{1}{3\sqrt{1}+3\sqrt{3}}+\dfrac{1}{3\sqrt{3}+3\sqrt{5}}+...+\dfrac{1}{3\sqrt{2017}+3\sqrt{2019}}\)
b ) S = \(\dfrac{1}{\sqrt{2.2}+\sqrt{2.3}}+\dfrac{1}{\sqrt{2.3}+\sqrt{2.4}}+\dfrac{1}{\sqrt{2.4}+\sqrt{2.5}}+...+\dfrac{1}{\sqrt{2.2018}+\sqrt{2.2019}}\)
Câu 1:
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
\(\Leftrightarrow1-\dfrac{1}{n+1}=\dfrac{2999}{3000}\)
=>n+1=3000
hay n=2999
