Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{58}\right)⋮7\)
a: \(G=8^8+2^{20}\)
\(=2^{24}+2^{20}\)
\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)
b: Sửa đề: \(H=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
\(H=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)⋮7\)
\(H=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{57}\right)⋮15\)
c: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)
\(E=1+3+3^2+3^3+...+3^{1991}\)
\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)
\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{97}\right)\)
\(=30\left(1+2^4+...+2^{96}\right)⋮30\)
2:
\(B=3+3^2+3^3+...+3^{2022}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)
\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)
\(A=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+...+3^{58}\right)\\ A=13\left(3+...+3^{58}\right)⋮13\)
\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\\ M=\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\\ M=\left(2+2^2+2^3+2^4\right)\left(1+...+2^{16}\right)\\ M=30\left(1+...+2^{16}\right)⋮5\)
a: \(A=\left(1+3\right)+...+3^{10}\left(1+3\right)\)
\(=4\left(1+...+3^{10}\right)⋮4\)
Bài 1
a, cm : A = 165 + 215 ⋮ 3
A = 165 + 215
A = (24)5 + 215
A = 220 + 215
A = 215.(25 + 1)
A = 215. 33 ⋮ 3 (đpcm)
b,cm : B = 88 + 220 ⋮ 17
B = (23)8 + 220
B = 216 + 220
B = 216.(1 + 24)
B = 216. 17 ⋮ 17 (đpcm)
c, cm: C = 1 - 2 + 22 - 23 + 24 - 25 + 26 -...-22021 + 22022 : 6 dư 1
C=1+(-2+22-23+24- 25+26)+...+(-22017+22018-22019+22020-22021+22022)
C = 1 + 42 +...+ 22016.(-2 + 22 - 23 + 24 - 25 + 26)
C = 1 + 42+...+ 22016.42
C = 1 + 42.(20+...+22016)
42 ⋮ 6 ⇒ C = 1 + 42.(20+...+22016) : 6 dư 1 đpcm
A=2+22+23+...+299+2100A=2+22+23+...+299+2100
⇒2A=22+23+24+...+2100+2101⇒2A=22+23+24+...+2100+2101
⇒A=2101−2⇒A=2101−2
B=3+32+33+...+399+3100B=3+32+33+...+399+3100
⇒3B=32+33+34+...+3100+3101⇒3B=32+33+34+...+3100+3101
⇒2B=3101−3⇒2B=3101−3
⇒B=3101−32
cho A = 1 + 3 + 32 + 33 + ... + 311
a ) chứng minh A chia hết cho 13
b) chứng minh A chia hết cho 40
A=1+3+3^2+3^3+...+3^98+3^99+3^100
A=(1+3+ 3^2)+(3^3+3^4+3^5)+...+(3^98+3^99+3^100)
A=(1+3+3^2)+3^3x(1+3+3^2)+...+3^98x(1+3+3^2)
A=13x3^3x13+...+3^98x13
=> 13x(1+3+3^3+...+3^98)chia hết cho 13
Vậy A chia hết cho 13
a) A= (1 + 3 + 32) + ( 33 + 34 + 35) + ... + (39 + 310 + 311)
= (1 + 3 + 32) + 32(1 + 3 + 32) + ... + 39(1 + 3 + 32)
= (1 + 3 + 32)(1 + 32 + ... + 39)
= 13(1 + 32 + ... + 39) chia hết 13
b) Tương tự, nhóm các số vào nhau như câu a) nhưng là 4 số một lần nhóm
c) Có 5n = ...5
47102 = (474)25 . 472 = (...1) . (...9) = ...9
=> 5n + 47102 = (...5) + (...9) = ...4 ko chia hết cho 10
=> đề sai :P