Question

Tính tổng

\(E=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\cdots+\frac{1}{3^{50}}+\frac{1}{3^{51}}\)

Answer 1

Sửa đề: \(E=-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^4}-...+\dfrac{1}{3^{50}}-\dfrac{1}{3^{51}}\)

=>\(3E=-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{49}}-\dfrac{1}{3^{50}}\)
=>\(3E+E=-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{49}}-\dfrac{1}{3^{50}}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{50}}-\dfrac{1}{3^{51}}\)

=>\(4E=-1-\dfrac{1}{3^{51}}=\dfrac{-3^{51}-1}{3^{51}}\)

=>\(E=\dfrac{-3^{51}-1}{4\cdot3^{51}}\)