Question

[0o0^^^Nhi^^^0o0]

So sánh:

a, \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{50}}\) và \(\dfrac{1}{2}\)

b, \(\dfrac{1}{4}-\dfrac{1}{4^2}+\dfrac{1}{4^3}-...+\dfrac{1}{4^{99}}\) và \(\dfrac{1}{12}\)

c, \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{50}{3^{50}}\) và \(\dfrac{3}{4}\)

Answer 1

[Nguyễn Thanh Hằng]

a/ Đặt :

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+.........+\dfrac{1}{3^{50}}\)

\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.......+\dfrac{1}{3^{49}}\)

\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+....+\dfrac{1}{3^{49}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{50}}\right)\)

\(\Leftrightarrow2A=1-\dfrac{1}{3^{50}}\)

còn sao nx thì mk chịu =.=