tìm số dư khi chia tổng sau cho 7
21 + 22+ 23 +...+298+ 299 + 2100
Tính số dư khi chia:
( 2 1 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 ) cho 7
Ta có
2 1 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 +...+ 2 98 + 2 99 + 2 100
= 2 1 + ( 2 2 + 2 3 + 2 4 ) + ( 2 5 + 2 6 + 2 7 ) +...+ ( 2 98 + 2 99 + 2 100 )
= 2 + 2 2 1 + 2 + 2 2 + 2 5 1 + 2 + 2 2 + . . . + 2 98 1 + 2 + 2 2
= 2 + 2 2 . 7 + 2 5 . 7 + . . . + 2 98 . 7 = 2 + 7 2 2 + 2 5 + . . . + 2 98
Mà 7 . 2 2 + 2 5 + . . . + 2 98 ⋮ 7
Nên 2 + 7 2 2 + 2 5 + . . . + 2 98 : 7 d ư 2
Tính số dư khi chia:
( 2 1 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100 ) cho 7
Số dư của A = 20 + 21 + 22 + 23 + .......... + 2100 khi chia cho 100 là: ................
Ta có A=20+21+22+23+...2100
2A=21+22+...+2101
2A-A=(21+22+...+2100)-(20+21+...+2100)
A=2101-1
Mà 2101-1=(........02)-1=........01 chia 100 dư 1
Chúc bạn học tốt.
CMR :
2100 - 299 + 298 - 297 + ...... + 24 - 23 + 22 ⋮ 12
\(2^{100}-2^{99}+2^{98}-2^{97}+2^{96}-2^{95}+...+2^4-2^3+2^2\)
\(=\left(2^{100}-2^{99}+2^{98}\right)-\left(2^{97}-2^{96}+2^{95}\right)+...+\left(2^4-2^3+2^2\right)\)
\(=2^{96}\left(2^4-2^3+2^2\right)-2^{93}\left(2^4-2^3+2^2\right)+...+\left(2^4-2^3+2^2\right)\)
\(=12\left(2^{96}-2^{93}+...+1\right)⋮12\)
A=2100-299+298-297+...-23+22-2+1
HELP ME
\(A=2^{100}-2^{99}+2^{98}-2^{97}+....-2^3+2^2-2+1\\ A=\left(2^{100}+2^{98}+...+2\right)-\left(2^{99}+2^{97}+...+1\right)\)
Gọi \(\left(2^{100}+2^{98}+...+2\right)\)là B
\(B=\left(2^{100}+2^{98}+...+2\right)\\ 2B=2^{102}+2^{100}+.....+2^2\\ 2B-B=\left(2^{102}+2^{100}+.....+2^2\right)-\left(2^{100}+2^{98}+...+2\right)\\ B=2^{102}-2\)
Gọi \(\left(2^{99}+2^{97}+...+1\right)\) là C
\(C=\left(2^{99}+2^{97}+...+1\right)\\ 2C=2^{101}+2^{99}+....+2\\ 2C-C=\left(2^{101}+2^{99}+9^{97}+...+2\right)-\left(2^{99}+9^{97}+...+1\right)\\ C=2^{101}-1\)
\(A=B+C\\ =>A=2^{102}-2+2^{101}-1\\ A=2^{101}\left(2+1\right)-3\\ A=2^{101}\cdot3-3\\ A=3\cdot\left(2^{101}-1\right)\)
\(\dfrac{1}{2}A=2^{99}-2^{98}+...-1+\dfrac{1}{2}\\ \Rightarrow A-\dfrac{1}{2}A=2^{100}-\dfrac{1}{2}\\ \Rightarrow A=2^{101}-1\)
Tìm số dư khi chia
1 + 2 + 22+ 23+ ... + 2100 chia cho 3
\(A=1+2+2^2+2^3+...+2^{100}\)
\(=1+\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\)
\(=1+2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(=1+3\left(2+2^3+...+2^{99}\right)\)
=>A chia 3 dư 1
thu gọn tổng sau
A= 2+22+23+24+...+299+2100
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(\Rightarrow2A=2^2+2^3+2^4+...+2^{100}+2^{101}\)
\(\Rightarrow A=2A-A=2^2+2^3+2^4+...+2^{100}+2^{101}-2-2^2-2^3-2^4-...-2^{99}-2^{100}=2^{101}-2\)
a, A = 1 + 2 + 22 + 23 + ... + 250 =
b, B = 1 + 3 + 32 + 33 + ... 3100 =
c, C = 5 + 52 + 53 + ... 530 =
d, D = 2100 = 299 + 298 - 297 + ... + 22 - 2
a) \(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)
b) \(B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+...+3^{101}\)
\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)
\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)
c) \(C=5+5^2+...+5^{30}\)
\(\Rightarrow5C=5^2+5^3+...+5^{31}\)
\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)
\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)
d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)
\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)
\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)
Tính:
A=2100-299-298-...-22-2-1
Ta có: \(A=2^{100}-2^{99}-2^{98}-...-2^2-2-1\)
\(\Leftrightarrow2A=2^{101}-2^{100}-2^{99}-...-2^3-2^2-2\)
\(\Leftrightarrow2A-A=2^{101}-2^{100}-2^{99}-...-2^3-2^2-2-2^{100}+2^{99}+2^{98}+...+2^2+2+1\)
\(\Leftrightarrow A=2^{101}-2\cdot2^{100}+1\)
\(\Leftrightarrow A=1\)
Rút gọn mỗi biểu thức sau :
câu a) A=1+3+3^2+3^3+.........................+3^99 +3^100
câu b) B=2100-299+298-297+....-23+22-2+1
mấy cái (/ ) là luỹ thừa
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