Toán Lớp 6: Chứng minh rằng :C=2+2^2+2^3+-.+2^99+2^100 chia hết cho 31

Question

Toán Lớp 6:  Chứng minh rằng :C=2+2^2+2^3+.....+2^99+2^100 chia hết cho 31

bichquyenlien · Answer

Answer    C = 2 + 2^{2} + 2^{3} + 2^{4} + 2^{5} + ... +2^{99} + 2^{100}   C = (2 + 2^{2} + 2^{3} + 2^{4} + 2^{5}) + ... + (2^{96} + 2^{97} + 2^{98} 2^{99} + 2^{100})   C = 2 . (1 + 2 + 2^{2} + 2^{3} + 2^{4}) + ... + 2^{96} . (1 + 2 + 2^{2} + 2^{3} + 2^{4})   C = 2 . 31 + ... + 2^{96} . 31   C = 31 . (2 + ... + 2^{96})   $ text{V&igrave;}$ 31  vdots 31 $ text{n&ecirc;n}$ 31 . (2 + ... + 2^{96})  vdots 31   => C  vdots 31   $ text{Vậy b&agrave;i to&aacute;n được chứng minh}$

thienthanh · Answer

C=2+2^2+2^3+2^4+2^5+...+2^96+2^97+2^98+2^99+2^100   C=(2+2^2+2^3+2^4+2^5)+...+(2^96+2^97+2^98+2^99+2^100)   C=2(1+2+2^2+2^3+2^4)+...+2^96(1+2+2^3+2^4+2^5)   C=2.31+...+2^96 .31   C=(2+...+2^96).31  vdots 31   =>C=2+2^2+2^3+2^4+2^5+...+2^96+2^97+2^98+2^99+2^100  vdots 31