Toán Lớp 6: Cho A= 2+2^2+2^3+2^4+-.+2^59+2^60 a) Tính A b) Chứng minh rằng A chia hết cho 21

Question

Toán Lớp 6:  Cho A= 2+2^2+2^3+2^4+.....+2^59+2^60 a) Tính A  b) Chứng minh rằng A chia hết cho 21

cobexinhdep · Answer

a, 2 + 2^2 + 2^3 +...+ 2^60     2A = 2(2 + 2^2 + 2^3 +...+ 2^60)     2A = 2^2 + 2^3 + 2^4 +...+ 2^61     2A - A = (2^2 + 2^3 + 2^4 +...+ 2^61) - (2 + 2^2 + 2^3 +...+ 2^60)     A = 2^2 + 2^3 + 2^4 +...+ 2^61 - 2 - 2^2 - 2^3 - 2^60     A = (2^2 - 2^2) + (2^3 - 2^3) +...+ (2^60 - 2^60) + 2^61 - 2     A = 2^61 - 2   b, A = 2 + 2^2 + 2^3 +...+ 2^60     = (2 + 2^2) + (2^3 + 2^4) +...+ (2^59 + 2^60)     = 2(1 + 2) + 2^3(1 + 2) +...+ 2^59(1 + 2)     = 2 &middot; 3 + 2^3 &middot; 3 +...+ 2^59 &middot; 3     = 3(2 + 2^3 + 2^5 +...+ 2^59) vdots 3     => A vdots 3  (1)     A = 2 + 2^2 + 2^3 +...+ 2^60     = (2 + 2^2 + 2^3) + (2^4 + 2^5 + 2^6) +...+ (2^58 + 2^59 + 2^60)     = 2(1 + 2 + 2^2) + 2^4(1 + 2 + 2^2) +...+ 2^58(1 + 2 + 2^2)     = 2 &middot; 7 + 2^4 &middot; 7 +...+ 2^58 &middot; 7     = 7(2 + 2^4 + 2^7 +...+ 2^58) vdots 7     => A vdots 7  (2)   Từ (1) v&agrave; (2)   => A vdots 21 $ text{(đpcm)}$

ngoclankhue · Answer

Answer    a, A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^59 + 2^60   2A = 2 . (2 + 2^2 + 2^3 + 2^4 + ... + 2^59 + 2^60)   2A = 2^2 + 2^3 + 2^4 + 2^5 + ... + 2^60 + 2^61   2A - A = (2^2 + 2^3 + 2^4 + 2^5 + ... + 2^60 + 2^61) - (2 + 2^2 + 2^3 + 2^4 + ... + 2^59 + 2^60)   A = 2^61 - 2   _____________________________   b, A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^59 + 2^60   A = (2 + 2^3 + 2^5) + (2^2 + 2^4 + 2^6) + ... + (2^55 + 2^57 + 2^59) + (2^56 + 2^58 + 2^60)   A = 2 . (1 + 2^2 + 2^4) + 2^2 . (1 + 2^2 + 2^4) + ... + 2^55 . (1 + 2^2 + 2^3) + 2^56 . (1 + 2^2 + 2^3)   A = 2 . 21 + 2^2 . 21 + ... + 2^55 . 21 + 2^56 . 21   A = 21 . (2 + 2^2 + ... + 2^55 + 2^56)  vdots 21   $ text{Vậy b&agrave;i to&aacute;n được chứng minh}$