Question

Tính:

\(B=-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\)\(\frac{1}{3^{50}}-\frac{1}{3^{51}}\)

Answer 1

\(B=\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{50}}-\frac{1}{3^{51}}\)

\(3B=-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{49}}-\frac{1}{3^{50}}\)

\(3B+B=\left(-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{49}}+\frac{1}{3^{50}}\right)\)

                      \(+\left(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{50}}-\frac{1}{3^{51}}\right)\)

\(4B=-1-\frac{1}{3^{51}}\)

\(B=\left(-1-\frac{1}{3^{51}}\right):4\)

\(B=\frac{-1}{4}\)

Answer 2

Mình cho bạn 1 công thức rồi tự làm 1 mình nhé:

Answer 3

\(B=\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{50}}-\frac{1}{3^{51}}\)

\(3B=-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}...+\frac{1}{3^{49}}+\frac{1}{3^{50}}\)

\(3B+B=\left(-1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{49}}+\frac{1}{3^{50}}\right)\)

\(+\left(\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{50}}-\frac{1}{3^{51}}\right)\)

\(4B=-1-\frac{1}{3^{51}}\)

\(B=\left(-1-\frac{1}{3^{51}}\right):4\)

\(B=\frac{-1}{4}\)