Toán S=2+2^2+2^3+2^4+2^5+…+2^98+2^99 chứng minh biểu thức S chia hết cho 14 23/07/2021 By Jasmine S=2+2^2+2^3+2^4+2^5+…+2^98+2^99 chứng minh biểu thức S chia hết cho 14
S= 2+2²+2³+…+$2^{99}$ = ( 2+2²+2³)+…+( $2^{97}$+$2^{98}$$2^{99}$) = 14+…+$2^{96}$.( 2+2²+2³) = 14+…+$2^{96}$.14 = 14.( 1+…+$2^{96}$)⋮ 14 Vậy S⋮ 14 Trả lời
S=2+2^2+2^3+2^4+2^5+…+2^98+2^99 =(2+2^2+2^3)+(2^4+2^5+2^6+2^7)+…+(2^96+2^97+2^98+2^99) =(2+2^2+2^3)+2^4(2+2^2+2^3)+…+2^96(2+2^2+2^3) =(2+2^2+2^3)(1+2^4+….+2^96) =14.(1+2^4+….+2^96) ⋮ 14 =>S ⋮ 14 Vậy biểu thức S chia hết cho 14. Trả lời
S= 2+2²+2³+…+$2^{99}$
= ( 2+2²+2³)+…+( $2^{97}$+$2^{98}$$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.