Question

Rút gọn:

A= 2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ... + 2² - 2

B= \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

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Answer 1

\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

\(2A+A=2^{101}-2\)

\(A=\frac{2^{101}-2}{3}\)

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(3B-B=1-\frac{1}{3^{99}}\)

\(B=\frac{1-\frac{1}{3^{99}}}{2}\)

Answer 2

\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

\(2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)

\(3A=2^{101}-2\)

\(A=\frac{2^{101}-2}{3}\)

Chúc bạn học tốt ~ 

Answer 3

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(2B=1-\frac{1}{3^{99}}\)

\(B=\frac{1-\frac{1}{3^{99}}}{2}\)

Chúc bạn học tốt ~