Question

[Jimin]

Thuc hien phep tinh:

B=1/3+1/3^2+1/3^3+...+1/3^2004+1/3^2005

Answer 1

[Hắc Hường]

Giải:

\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\)

\(\Leftrightarrow\dfrac{1}{3}B=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}...+\dfrac{1}{3^{2005}}+\dfrac{1}{3^{2006}}\)

\(\Leftrightarrow B-\dfrac{1}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2006}}\)

\(\Leftrightarrow\dfrac{2}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2006}}\)

\(\Leftrightarrow B=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^{2006}}}{\dfrac{2}{3}}\)

\(\Leftrightarrow B=\dfrac{1-\dfrac{1}{3^{2005}}}{2}\)

\(\Leftrightarrow B=\dfrac{\dfrac{3^{2005}-1}{3^{2005}}}{2}\)

\(\Leftrightarrow B=\dfrac{3^{2005}-1}{2.3^{2005}}\)

Vậy ...

Answer 2

[Tram Nguyen]

Chúc bn học tốt!