Question

[Ngọc Hân Cao Dương]

A= 3¹⁰⁰- 3⁹⁹+ 3⁹⁸-...+ 3²- 3
B= (-2)⁰+ (-2)¹+ (-2)²+...+ (-2)²⁰²⁴
C= (\(\dfrac{-1}{5}\))⁰+ (\(\dfrac{-1}{5}\))¹+ (\(\dfrac{-1}{5}\))²+....+ (\(\dfrac{-1}{5}\))²⁰²³

Answer 1

[Nguyễn Lê Phước Thịnh]

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

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

=>\(4A=3^{101}-3\)

=>\(A=\dfrac{3^{101}-3}{4}\)

b: \(B=\left(-2\right)^0+\left(-2\right)^1+...+\left(-2\right)^{2024}\)

=>\(B\cdot\left(-2\right)=\left(-2\right)^1+\left(-2\right)^2+...+\left(-2\right)^{2025}\)

=>\(-2B-B=\left(-2\right)^1+\left(-2\right)^2+...+\left(-2\right)^{2025}-\left(-2\right)^0-\left(-2\right)^1-...-\left(-2\right)^{2024}\)

=>\(-3B=-2^{2025}-1\)

=>\(B=\dfrac{2^{2025}+1}{3}\)

c: \(C=\left(-\dfrac{1}{5}\right)^0+\left(-\dfrac{1}{5}\right)^1+...+\left(-\dfrac{1}{5}\right)^{2023}\)

=>\(\left(-\dfrac{1}{5}\right)\cdot C=\left(-\dfrac{1}{5}\right)^1+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{2024}\)

=>\(\left(-\dfrac{6}{5}\right)\cdot C=\left(-\dfrac{1}{5}\right)^{2024}-\left(-\dfrac{1}{5}\right)^0\)

=>\(C\cdot\dfrac{-6}{5}=\dfrac{1}{5^{2024}}-1=\dfrac{1-5^{2024}}{5^{2024}}\)

=>\(C\cdot\dfrac{6}{5}=\dfrac{5^{2024}-1}{5^{2024}}\)

=>\(C=\dfrac{5^{2024}-1}{5^{2024}}:\dfrac{6}{5}=\dfrac{5^{2024}-1}{6\cdot5^{2023}}\)