Question

Rút gọn biểu thức:

a) A=1+3+\(^{ }3^2\)+...+\(3^{200}\)

b)B=4+4+\(4^2\)+4+...+\(^{ }4^{200}\)

Answer 1

a) Ta có: A = 1 + 3 + + ... +

⇒ 3A = 3 + + ... +

⇒ 3A - A = (3 + + ... + 1 + 3 + + ... +

⇒ 2A = 3²⁰¹ - 1

⇒ A = 2A : A = \(\dfrac{3^{201}-1}{2}\)

Vậy A = \(\dfrac{3^{201}-1}{3}\)

Answer 2

a: \(3A=3+3^2+3^3+...+3^{201}\)

=>\(2A=3^{201}-1\)

hay \(A=\dfrac{3^{201}-1}{2}\)

b: \(B=4+4^2+4^3+...+4^{200}\)

=>\(4B=4^2+4^3+4^4+...+4^{201}\)

=>\(3B=4^{201}-4\)

hay \(B=\dfrac{4^{201}-4}{3}\)