Question

Rút gọn biêủ thức :

A = 1 + 3 + \(3^2\) + \(3^3\) + ... + \(3^{39}\) + \(3^{100}\)

B = 1 - 3 + \(3^2\) + \(3^3\) + ... + \(3^{99}\) + \(3^{100}\)

Answer 1

Giải:

a) \(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)

\(\Leftrightarrow3A=3+3^2+3^3+3^4+...+3^{100}+3^{101}\)

\(\Leftrightarrow3A-A=2A=3^{101}-1\)

\(\Leftrightarrow A=\dfrac{3^{101}-1}{2}\)

b) \(B=1-3+3^2+3^3+...+3^{99}+3^{100}\)

\(\Leftrightarrow3B=3-3^2+3^3+3^4+...+3^{100}+3^{101}\)

\(\Leftrightarrow3B-B=2B=3^{101}-1-6-18=3^{101}--25\)

\(\Leftrightarrow B=\dfrac{3^{101}-25}{2}\)

Chúc bạn học tốt!

Answer 2

\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)

\(3A=3+3^2+3^3+3^4+...+3^{100}+3^{101}\)

\(3A-A=\left(3+3^2+3^3+3^4+...+3^{100}+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{99}\right)\)

\(2A=3^{101}-1\Leftrightarrow A=\dfrac{3^{101}-1}{2}\)

B đề sai