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a, A = 2 + 22 + 23 + 24 +....+ 260
A = (2 + 22) + ( 23 + 24) +...+ (259 + 260)
A = 2.(1 + 2) + 23.(1 + 2) +...+ 259.(1 + 2)
A = 2.3 + 23.3 +...+ 259.3
A = 3.( 2 + 23+...+ 259) vì 3 ⋮ 3 ⇒ A = 3.(2 + 23 +...+ 259) ⋮ 3 (đpcm)
A = 2 + 22 + 23+ 24+...+ 260
A = ( 2 + 22 + 23) + ( 24 + 25 + 26) +...+ (258 + 259 + 260)
A = 2.( 1 + 2 + 4) + 24.(1 + 2 + 4)+...+ 258.(1 + 2+4)
A = 2.7 + 24.7 +...+258.7
A = 7.(2 + 24 + ...+ 258) vì 7 ⋮ 7 ⇒ A = 7.(2 + 24+...+ 258)⋮ 7(đpcm)
A = 2 + 22 + 23 + 24 +...+ 260
A = (2 + 22 + 23 + 24) +...+( 257 + 258 + 259+ 260)
A = 2.(1 + 2 + 22 + 23) +...+ 257.(1 + 2 + 22+23)
A = 2.30 + ...+ 257. 30
A = 30.( 2 +...+ 257) vì 30 ⋮ 15 ⇒ 30.( 2 + ...+ 257) ⋮ 15 (đpcm)
a)
- \(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{59}.3\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
- \(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=2.7+2^4.7+...+2^{58}.7\)
\(=7\left(2+2^4+2^{58}\right)⋮7\)
- \(A=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=2.15+2^5.15+...+2^{57}.15\)
\(=15\left(2+2^5+2^{57}\right)⋮15\)
b) \(B=1+5+5^2+5^3+...+5^{96}+5^{97}+5^{98}\)
\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{96}+5^{97}+5^{98}\right)\)
\(=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+..+5^{96}\left(1+5+5^2\right)\)
\(=31+5^3.31+...+5^{96}.31\)
\(=31\left(1+5^3+...+5^{96}\right)⋮31\)
Chứng tỏ rằng :
a) 1+5+52+53+.......+5501 \(⋮\)6
b) 2+22 +23 +.. + 2100 vừa \(⋮\)31, vừa \(⋮\) cho 5
a/ \(1+5+5^2+..........+5^{501}\)
\(=\left(1+5\right)+\left(5^2+5^3\right)+............+\left(5^{500}+5^{501}\right)\)
\(=1\left(1+5\right)+5^2\left(1+5\right)+...........+5^{500}\left(1+5\right)\)
\(=1.6+5^2.6+.............+5^{500}.6\)
\(=6\left(1+5^2+..........+5^{500}\right)⋮6\left(đpcm\right)\)
b/ \(2+2^2+2^3+............+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+............+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+............+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+..........+2^{96}.31\)
\(=31\left(2+........+2^{96}\right)⋮31\left(đpcm\right)\)
a)1+5+5^2+5^3+........+5^501
= 6+(5^2+5^3)+(5^4+5^5)......+(5^500+5^501)
=6+150+150(5^2+5^3)+150(5^4+5^5).......150(5^499+5^500)
=6+150(5^2+5^3+.......+5^500)
mà 6 chia hết cho 6
150(5^2+5^3+.......+5^500) chia hết cho 6
=> 6+150(5^2+5^3+.......+5^500) chia hết cho 6
=> 6+150+150(5^2+5^3)+150(5^4+5^5).......150(5^499+5^500) chia hết cho 6
=> 6+(5^2+5^3)+(5^4+5^5)......+(5^500+5^501) chia hết cho 6
=> 1+5+5^2+5^3+........+5^501 chia hết cho 6
câu hỏi này nhiều rồi nhé: xem câu hỏi tương tự
Nếu không muốn: Gợi ý : cứ nhóm 2 số hạng một Ví dụ (5 + 5^2) = 5(5+1) = 5.6......
các nhóm đều xuất hiện thừa số 6 nên tổng chia hết cho 6.....
câu b cũng tương tự
\(\left(2^{10}+2^9\right)+\left(2^8+2^7\right)+....+\left(2^2+2\right)\)
\(=2^9.\left(2+1\right)+2^7.\left(2+1\right)+...+2.\left(2+1\right)\)
\(=2^9.3+2^7.3+...+2.3\)
\(=3.\left(2^9+2^7+...+2\right)⋮3\)
P/S: mấy bài khác tương tự
\(a,2^{10}+2^9+2^8+...+2\)
\(=\left(2^{10}+2^9\right)+\left(2^8+2^7\right)+...+\left(2^2+2\right)\)
\(=2^9\left(2+1\right)+2^7\left(2+1\right)+...+2\left(2+1\right)\)
\(=2^9.3+2^7.3+...+2.3\)
\(=3\left(2^9+2^7+...+2\right)⋮3\left(đpcm\right)\)
\(b,1+3+3^2+3^3+...+3^{99}\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)\)
\(=4+3^2\left(1+3\right)+...+3^{98}\left(1+3\right)\)
\(=4+3^2.4+...+3^{98}.4\)
\(=4\left(1+3^2+...+3^{98}\right)⋮4\left(đpcm\right)\)
\(c,1+5+5^2+5^3+...+5^{1975}\)
\(=\left(1+5\right)+\left(5^2+5^3\right)+...+\left(5^{1974}+5^{1975}\right)\)
\(=6+5^2\left(1+5\right)+...+5^{1974}\left(1+5\right)\)
\(=6+5^2.6+...+5^{1974}.6\)
\(=6\left(1+5^2+...+5^{1974}\right)⋮6\left(đpcm\right)\)