Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
S = 3 + 3 2 + 3 3 + 3 4 + 3 5 + 3 6 + 3 7 + 3 8 + 3 9 = 3 + 3 2 + 3 3 + 3 4 + 3 5 + 3 6 + 3 7 + 3 8 + 3 9 = 39 + 3 3 . 39 + 3 6 . 39 = 39 . 1 + 3 3 + 3 6 ⋮ − 39
Vậy S chia hết cho -39
S = 3 + 3 2 + 3 3 + 3 4 + 3 5 + 3 6 + 3 7 + 3 8 + 3 9 = 3 + 3 2 + 3 3 + 3 4 + 3 5 + 3 6 + 3 7 + 3 8 + 3 9 = 39 + 3 3 . 39 + 3 6 . 39 = 39. 1 + 3 3 + 3 6 ⋮ − 39
Vậy S chia hết cho -39
\(A,\)\(S=\left(3+3^2\right)+\left(3+3^2\right)3^2+...+\left(3+3^2\right)3^{2018} \)
\(\Rightarrow S=9\left(1+3^2+...+3^{2018}\right)\)
\(\Rightarrow S⋮9\)
\(B,\)\(S=3+3^2+3^3+\left(3+3^2+3^3\right)3^3+...\left(3+3^2+3^3\right)3^{2017}\)
\(S=39+39.3^3+...+39.3^{2017}\)
Nhưng xét lại thì thấy 2017 không chia hết cho 3 nên câu b có lẽ sai đề =)))))
\(C,\)\(S=\left(1+3+3^2+3^3\right).3+\left(1+3+3^2+3^3\right).3^4+...+\left(1+3+3^2+3^3\right).3^{2017}\)
\(S=40.3+40.3^4+...+40.3^{2017}\)
\(Vậy...\)
Bài 1:
\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)
Bài 2:
\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)
Bài 1 :
\(2^{49}=\left(2^7\right)^7=128^7\)
\(5^{21}=\left(5^3\right)^7=125^7\)
mà \(125^7< 128^7\)
\(\Rightarrow2^{49}>5^{21}\)
Bài 2 :
a) \(S=1+3+3^2+3^3+...3^{99}\)
\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)
\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)
\(\Rightarrow dpcm\)
b) \(S=1+4+4^2+4^3+...4^{62}\)
\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)
\(\Rightarrow S=21+4^3.21+...4^{60}.21\)
\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)
\(\Rightarrow dpcm\)
\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)
\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)
\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)
\(S=4\left(3^2+3^4+3^6+3^8\right)\)
\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4
\(S=3+3^2+3^3+...+3^{1998}\)
\(S=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{1997}+3^{1998}\right)\)
\(S=12+3^2\cdot\left(3+3^2\right)+...+3^{1996}\cdot\left(3+3^2\right)\)
\(S=12\cdot1+12\cdot3^2+...+12\cdot3^{1996}\)
\(S=12\cdot\left(1+3^2+...+3^{1996}\right)⋮12\)
b, tương tự nhưng nhóm 3 số hạng
Bài ở đâu đấy Ly, k cho tớ đi!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!