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Ta có : 4n+1 +2.4n = 384
<=> 4n.4 + 2.4n = 384
<=> 4n.(4 + 2) = 384
<=> 4n.6 = 384
<=> 4n = 64
<=> 4n = 43
<=> n = 3
@Thảo Nguyên
Bài 1:
\(A=\dfrac{2}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{89.93}\)
\(A=\dfrac{2}{1.5}+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{89}-\dfrac{1}{93}\right)\)
\(A=\dfrac{2}{5}+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{93}\right)\)
\(A=\dfrac{2}{5}+\dfrac{1}{4}.\dfrac{88}{465}\)
\(A=\dfrac{2}{5}+\dfrac{22}{465}=\dfrac{208}{465}\)
1. Mk sửa lại đề bài như sau:
\(A=\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{89.93}\)
\(\Rightarrow4A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{89.93}\)
\(4A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{89}-\dfrac{1}{93}\)
\(4A=1-\dfrac{1}{93}\)
\(4A=\dfrac{92}{93}\)
\(A=\dfrac{92}{93}:4\)
\(A=\dfrac{23}{93}\)
2. Mk cux sửa lại đề bài:
\(A=3+3^2+3^3+3^4+3^5+...+3^{100}\)
\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=3\left(1+3+9+27\right)+...+3^{97}\left(1+3+9+27\right)\)
\(=3.40+...+3^{97}.40\)
\(=\left(3+3^{97}\right)⋮4.10\)
\(\Rightarrow A⋮4;10\)
b)\(2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+....+2^{96}.31\)
\(=31.\left(2+....+2^{96}\right)⋮31\)
Vậy...
a) \(5+5^2+5^3+...+5^{2004}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...\left(5^{2003}+5^{2004}\right)\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{2003}\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{2003}.6\)
\(=6.\left(5+5^3+...+5^{2003}\right)⋮6\)
Vậy....
\(5+5^2+5^3+...+5^{2004}\)
\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6+\right)+...+\left(5^{2002}+5^{2003}+5^{2004}\right)\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{2002}\left(1+5+5^2\right)\)
\(=5.31+5^4.31+...+5^{2002}.31\)
\(=31.\left(5+5^4+...+5^{2002}\right)⋮31\)
Vậy...
Trường hợp 3 làm tương tự để chứng minh
a)
\(123⋮3\\ 7\cdot3\cdot11119⋮3\\ \Rightarrow123+7\cdot3\cdot11119⋮3\)
Vậy \(123+7\cdot3\cdot11119⋮3\)
c)
\(8^8+2^{20}=\left(2^3\right)^8+2^{20}=2^{24}+2^{20}=2^{20}\cdot\left(2^4+1\right)=2^{20}\cdot\left(16+1\right)=2^{20}\cdot17⋮17\)
Vậy \(8^8+2^{20}⋮17\)
d)
Ta thấy:
\(...2^4=...6,...2^8=...6\Rightarrow...2^{4n}=...6\left(n\in N^{\circledast}\right)\)
\(...1^n=...1\left(n\in N^{\circledast}\right)\)
\(\left(...2\text{ là số có chữ số tận cùng là }2,\text{ tương tự với }...1,...6,...5\: \right)\)
\(\Rightarrow942^{60}-351^{37}=942^{4\cdot15}-351^{37}=...6-...1=...5⋮5\)
Vậy \(942^{60}-351^{37}⋮5\)
1)
a) Gọi 2 số tự nhiên liên tiếp là a ; a+1
Ta chứng minh : a . (a+1) chia hết cho
=> a.a + a.1
=> 2a + a
Vì 2a chia hết cho 2
=> a chia hết cho 2
=> a . (a+1) chia hết cho 2 đpcm
vì sao 2a\(⋮\)2 thì a\(⋮\)2 hở bn? vd: 2a=6 thì sao