S=2+2^2+2^3+2^4+2^5+…+2^98+2^99 chứng minh biểu thức S chia hết cho 14 Question S=2+2^2+2^3+2^4+2^5+…+2^98+2^99 chứng minh biểu thức S chia hết cho 14 in progress 0 Toán Jasmine 1 năm 2021-07-23T08:36:46+00:00 2021-07-23T08:36:46+00:00 2 Answers 8 views 0
Answers ( )
S= 2+2²+2³+…+$2^{99}$
= ( 2+2²+2³)+…+( $2^{97}$+$2^{98}$$2^{99}$)
= 14+…+$2^{96}$.( 2+2²+2³)
= 14+…+$2^{96}$.14
= 14.( 1+…+$2^{96}$)⋮ 14
Vậy S⋮ 14
S=2+2^2+2^3+2^4+2^5+…+2^98+2^99
=(2+2^2+2^3)+(2^4+2^5+2^6+2^7)+…+(2^96+2^97+2^98+2^99)
=(2+2^2+2^3)+2^4(2+2^2+2^3)+…+2^96(2+2^2+2^3)
=(2+2^2+2^3)(1+2^4+….+2^96)
=14.(1+2^4+….+2^96) ⋮ 14
=>S ⋮ 14
Vậy biểu thức S chia hết cho 14.