Toán chứng mình rằng (2+2^2+2^3+2^4+…+2^8+2^9)/7 giúp em vói mai em nộp bài rồi 14/09/2021 By Brielle chứng mình rằng (2+2^2+2^3+2^4+…+2^8+2^9)/7 giúp em vói mai em nộp bài rồi
Đáp án: 2+2^2+2^3+2^4+…+2^8+2^9 = (2+2^2 + 2^3) + (2^4+2^5+2^6) + (2^7+2^8+2^9) = (2+2^2 + 2^3) + 2^3(2+ 2^2+2^3) + 2^6(2+2^2+2^3) = 14 . 1 + 2^3 . 14 + 2^6 . 14 ⇒ 14 ( 1+ 2^3 + 2^6) chia hết cho 7 ⇔ (2+2^2+2^3+2^4+…+2^8+2^9)/7 Trả lời
$2+2^{2}+2^{3}+2^{4}+…+2^{8}+2^{9}$ $= (2+2^{2} + 2^{3}) + (2^{4}+2^{5}+2^{6}) + (2^{7}+2^{8}+2^{9})$ $= (2+2^{2} + 2^{3}) + 2^{3}(2+ 2^{2}+2^{3}) + 2^{6}(2+2^{2}+2^{3})$ $= 14 . 1 + 2^{3} . 14 + 2^{6} . 14$ $⇒ 14 ( 1+ 2^3 + 2^6)$ $\vdots$ $7$ $→ (2+2^{2}+2^{3}+2^{4}+…+2^{8}+2^{9})$ $\vdots$ $7$ Trả lời
Đáp án:
2+2^2+2^3+2^4+…+2^8+2^9
= (2+2^2 + 2^3) + (2^4+2^5+2^6) + (2^7+2^8+2^9)
= (2+2^2 + 2^3) + 2^3(2+ 2^2+2^3) + 2^6(2+2^2+2^3)
= 14 . 1 + 2^3 . 14 + 2^6 . 14
⇒ 14 ( 1+ 2^3 + 2^6) chia hết cho 7
⇔ (2+2^2+2^3+2^4+…+2^8+2^9)/7
$2+2^{2}+2^{3}+2^{4}+…+2^{8}+2^{9}$
$= (2+2^{2} + 2^{3}) + (2^{4}+2^{5}+2^{6}) + (2^{7}+2^{8}+2^{9})$
$= (2+2^{2} + 2^{3}) + 2^{3}(2+ 2^{2}+2^{3}) + 2^{6}(2+2^{2}+2^{3})$
$= 14 . 1 + 2^{3} . 14 + 2^{6} . 14$
$⇒ 14 ( 1+ 2^3 + 2^6)$ $\vdots$ $7$
$→ (2+2^{2}+2^{3}+2^{4}+…+2^{8}+2^{9})$ $\vdots$ $7$