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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
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.
Lời giải:
Đặt $A=1+2^2+2^4+....+2^{100}$
$A=(1+2^2+2^4)+(2^6+2^8+2^{10})+.....+(2^{96}+2^{98}+2^{100})$
$A=(1+2^2+2^4)+2^6(1+2^2+2^4)+....+2^{96}(1+2^2+2^4)$
$=(1+2^2+2^4)(1+2^6+....+2^{96})$
$=21(1+2^6+....+2^{96})\vdots 21$
Ta có đpcm.
\(A=2^{100}-\left(2^{99}+2^{98}+...+2+1\right)\)
Đặt \(B=2^{99}+2^{98}+...+2+1\)
\(\Rightarrow2B=2^{100}+2^{99}+...+2^2+2\)
\(\Rightarrow2B-B=2^{100}-1\Leftrightarrow B=2^{100}-1\)
\(\Rightarrow A=2^{100}-\left(2^{100}-1\right)=1\)
`#3107.101107`
Gọi biểu thức trên là A
Ta có:
\(A=1+5^2+5^4+...+5^{40}\\ =1\cdot\left(1+5^2\right)+5^4\cdot\left(1+5^2\right)+...+5^{38}\cdot\left(1+5^2\right)\\ =\left(1+5^2\right)\cdot\left(1+5^4+...+5^{38}\right)\\ =26\cdot\left(1+5^4+...+5^{38}\right)\)
Vì \(26\cdot\left(1+5^4+...+5^{38}\right)\text{ }⋮\text{ }26\)
\(\Rightarrow A\text{ }⋮\text{ }26\)
_______
Gọi biểu thức trên là B
Ta có:
\(B=1+2^2+2^4+...+2^{100}\\ =1\cdot\left(1+2^2+2^4\right)+2^6\cdot\left(1+2^2+2^4\right)+...+2^{96}\cdot\left(1+2^2+2^4\right)\\ =\left(1+2^2+2^4\right)\cdot\left(1+2^6+...+2^{96}\right)\\ =21\cdot\left(1+2^6+...+2^{96}\right)\)
Vì \(21\cdot\left(1+2^6+...+2^{96}\right)\text{ }⋮\text{ }21\)
\(\Rightarrow B\text{ }⋮\text{ }21\)
_______
Gọi biểu thức trên là C
Ta có:
\(C=1+3^2+3^4+...+3^{100}\\ =1\cdot\left(1+3^2+3^4+3^6\right)+3^6\cdot\left(1+3^2+3^4+3^6\right)+...+3^{94}\cdot\left(1+3^2+3^4+3^6\right)\\ =\left(1+3^2+3^4+3^6\right)\cdot\left(1+3^6+...+3^{94}\right)\\ =820\cdot\left(1+3^6+...+3^{94}\right)\)
Vì \(820\cdot\left(1+3^6+...+3^{94}\right)\text{ }⋮\text{ }82\)
\(\Rightarrow C\text{ }⋮\text{ }82.\)
a) \(A=1+5^2+5^4+5^6...+5^{40}\)
\(\Rightarrow A=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{38}\left(1+5^2\right)\)
\(\Rightarrow A=26+5^4.26+...+5^{38}.26\)
\(\Rightarrow A=26\left(1+5^4+...+5^{38}\right)⋮26\)
\(\Rightarrow1+5^2+5^4+5^6...+5^{40}⋮6\left(dpcm\right)\)
b) \(B=1+2^2+2^4+2^6+...+2^{100}\)
\(\Rightarrow B=\left(1+2^2+2^4\right)+2^6\left(1+2^2+2^4\right)+...+2^{96}\left(1+2^2+2^4\right)\)
\(\Rightarrow B=21+2^6.21+...+2^{96}.21\)
\(\Rightarrow B=21\left(1+2^6+...+2^{96}\right)⋮21\)
\(\Rightarrow1+2^2+2^4+2^6+...+2^{100}⋮21\left(dpcm\right)\)
Bài C tương tự bạn tự làm nhé!