Question

Y=\(\dfrac{1}{^{^{^{ }}}3^0}+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2005}}\)

TÍNH TỔNG

Answer 1

Ta có :

\(Y=\dfrac{1}{3^0}+\dfrac{1}{3^1}+\dfrac{1}{3^2}+................+\dfrac{1}{3^{2005}}\)

\(\Rightarrow3Y=2+\dfrac{1}{3^0}+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...........+\dfrac{1}{3^{2004}}\)

\(\Rightarrow3Y-Y=\left(2+\dfrac{1}{3^0}+\dfrac{1}{3^1}+.............+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3^0}+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...........+\dfrac{1}{3^{2005}}\right)\)\(\Rightarrow2Y=2-\dfrac{1}{3^{2005}}\)

\(\Rightarrow Y=\dfrac{2-\dfrac{1}{3^{2005}}}{2}\)