Question

m = 4^{10 +} 4^{11 + ... +} 4¹⁹⁸ + 4 ¹⁹⁹ chung minh rang M la boi cua 5

 

Answer 1

Ta có:

\(M=4^{10}+4^{11}+...+4^{199}\)

\(\Rightarrow M=\left(4^{10}+4^{11}\right)+...+\left(4^{198}+4^{199}\right)\)

\(\Rightarrow M=4^{10}\left(1+4\right)+...+4^{198}\left(1+4\right)\)

\(\Rightarrow M=4^{10}.5+...+4^{198}.5\)

\(\Rightarrow M=\left(4^{10}+...+4^{198}\right).5⋮5\)

\(\Rightarrow M⋮5\)

\(\Rightarrow M\in B\left(5\right)\left(đpcm\right)\)

Answer 2

ai tra loi se co qua

 

Answer 3

Ta có: M = 4^10+4^11+ ...+4^199 (tổng m có 190 số hạng)

M=(4^10+4^11)+(4^12+4^13)+ ... +(4^198+4^199)

M=4^10(1+4)+4^12(1+4)+4^198(1+4)

M=4^10*5+4^12*5+ ... +4^198*5

M=5(4^10+4^12+...+4^198)

Vì 5(4^10+4^12+...+4^198) chia hết cho 5 nên M chia hết cho 5

 

 

Quà đâu bạn