Question

Cho biểu thức A=3¹+3²+3⁴+….+3⁶⁰.chứng tỏ rằng A chia hết cho 40.

nhanh giúp mk với ạ cảm ơn

 

Answer 1

\(A=3+3^2+3^3+...+3^{60}\)

\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{57}+3^{58}+3^{59}+3^{60}\right)\)

\(\Rightarrow A=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{57}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=\left(3+3^5+...+3^{57}\right)\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=40\left(3+3^5+...+3^{57}\right)⋮40\)

Answer 2

A=3+32+33+...+360

 

 

A=3+32+33+...+360⇒A=(3+32+33+34)+(35+36+37+38)+...+(357+358+359+360)⇒A=(3+32+33+34)+(35+36+37+38)+...+(357+358+359+360)

⇒A=3(1+3+32+33)+35(1