Question

A=3+3²+3³+....+3⁸⁹+3⁹⁰

a) Tính xem A có bao nhiêu số hạng.

b) Chứng minh rằng A chia hết cho 4, chia hết cho 13

Answer 1

A = \(3+3^2+3^3+.......+3^{89}+3^{90}\)

a) 

Số số hạng của A là :

(90 - 1) : 1 + 1 = 90 (số)

b)

A = \(3\left(1+3\right)+3^3\left(1+3\right)+3^5\left(1+3\right)+.......+3^{89}\left(1+3\right)\)

=> A = \(3\cdot4+3^3\cdot4+3^5\cdot4+.......+3^{89}\cdot4\)

=> A = \(\left(3+3^3+3^5+.....+3^{89}\right)\cdot4⋮4\)

A = \(3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^7\left(1+3+3^2\right)+.......+3^{87}\left(1+3+3^2\right)\)

=> A = \(13\left(3+3^4+3^7+......+3^{87}\right)⋮13\)

Answer 2

cảm ơn bạn đã giúp mình

Answer 3

\(A=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)

\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)