Cho : A = 1 + 2 + 22 + 23 + ..... + 2100 ; B = 2101 - 1 . so sánh A vs B
Thu gọn A và tìm n e N biết A + 2 = 2n
Bài 5: (1 điểm) Cho A= 2+22+23+24+.....+2100 . Chứng minh A chia hết cho 3.
Lời giải:
$A=(2+2^2)+(2^3+2^4)+....+(2^{99}+2^{100})$
$=2(1+2)+2^3(1+2)+...+2^{99}(1+2)$
$=2.3+2^3.3+...+2^{99}.3$
$=3(2+2^3+...+2^{99})\vdots 3$
Ta có đpcm.
Cho A = 2+22+23+24+...........+2100. Chứng minh A chia hết cho 3.
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
\(C=\left(2^{99}+2^{97}+...+1\right)\\ 2C=2^{101}+2^{99}+....+2\\ 2C-C=\left(2^{101}+2^{99}+9^{97}+...+2\right)-\left(2^{99}+9^{97}+...+1\right)\\ C=2^{101}-1\)
\(A=B+C\\ =>A=2^{102}-2+2^{101}-1\\ A=2^{101}\left(2+1\right)-3\\ A=2^{101}\cdot3-3\\ A=3\cdot\left(2^{101}-1\right)\)
\(\dfrac{1}{2}A=2^{99}-2^{98}+...-1+\dfrac{1}{2}\\ \Rightarrow A-\dfrac{1}{2}A=2^{100}-\dfrac{1}{2}\\ \Rightarrow A=2^{101}-1\)
Chứng tỏ rằng A = 2 + 22 + 23 + …+ 2100 chia hết cho 6.
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\\ A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\\ A=\left(2+2^2\right)\left(1+2^2+...+2^{98}\right)\\ A=6\left(1+2^2+...+2^{98}\right)⋮6\)
chứng tỏ A chia hết cho 6 với:
A=2+22+23+...+2100
\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6+2^2.6+...+2^{98}.6\)
\(=6\left(1+2^2+...+2^{98}\right)⋮6\)
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(=\left(2+2^2\right)+2^2\left(2+2\right)+...+2^{98}\left(2+2^2\right)\)
\(=\left(2+2^2\right)\left(1+2^2+...+2^{98}\right)\)
\(=6\left(1+2^2+...+2^{98}\right)\)⋮6
⇒ A⋮6
chứng tỏ A chia hết cho 6 với A= 2+22+23+24+...+2100
\(A=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6+2^2.6+...+2^{98}.6=6\left(1+2^2+...+2^{98}\right)⋮6\)
Chứng tỏ A chia hết cho 6 với A = 2 + 22+23+24+...+2100
\(A=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6+6.2^2+...+6.2^{98}\)
\(=6\left(1+2^2+...+2^{98}\right)⋮6\)
Chứng tỏ A chia hết cho 6 với A = 2 + 22 + 23 + 24 + … + 2100
\(A=2+2^2+2^3+2^4+...+2^{100}\)
\(=2\cdot3+2^3\cdot3+...+2^{99}\cdot3\)
\(=6\left(1+2^2+...+2^{98}\right)⋮6\)
Chứng tỏ A chia hết cho 10 với A = 2+22+23+...+2100 ❤
A=(2+2^2+2^3+2^4)+2^4(2+2^2+2^3+2^4)+...+2^96(2+2^2+2^3+2^4)
=30(1+2^4+...+2^96) chia hết cho 10