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Phần a:
Có 100 số tự nhiên chia làm 20 nhóm từ trái sang phải mỗi nhóm năm số.
\(C=2.\left(1+2+4+8+16\right)+2^6.\left(1+2+4+8+16\right)+...+2^{96}.\left(1+2+4+8+16\right)\)
\(C=2.31+2^6.31+2^{11}.31+...+2^{96}.31\)
=> C chia hết cho 31.
Chúc em học tốt^^
\(2.C=2^2+2^3+....+2^{101}\)
\(=>2C-C=C=2^2-2^2+2^3-2^3+....+2^{100}-2^{100}+2^{101}-2\)
\(C=2^{101}-2\)
Do đó 2x-1=101
=>x=51
Chúc em học tốt^^
a)\(C=2+2^2+2^3+....+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+2^5\left(2+2^2+2^3+2^4+2^5\right)+...+2^{95}\left(2+2^2+2^3+2^4+2^5\right)\)
\(=62+2^5.62+...+2^{95}.62=62\left(1+2^5+...+2^{95}\right)=31.2\left(1+2^5+....+2^{95}\right)⋮31\)
\(\Rightarrow C⋮31\)
=>đccm
\(C=2+2^2+2^3+...+2^{99}+2^{100}\)
\(C=\left(2+2^2+2^3+2^4+2^5\right)+....+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(C=2.\left(1+2+2^2+2^3+2^4\right)+....+2^{96}.\left(1+2+2^2+2^3+2^4\right)\)
\(C=31.2+.....+2^{96}.31=31.\left(2+....+2^{96}\right)⋮31\)
Suy ra \(C⋮31\)
b) Ta có \(2.C=2^2+2^3+2^4+....+2^{99}+2^{100}+2^{101}\)
Suy ra \(2.C-C=2^{101}-2\)hay \(C=2^{101}-2\)
Khi đó \(2^{2x-1}-2=2^{101}-2\)
\(\Rightarrow2^{2x-1}=2^{101}\)
\(\Rightarrow2x-1=101\Rightarrow2x=100\Rightarrow x=50\)
Vậy x = 50
Bài 1 : Ta có : \(A=3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10\left(3^n-2^{n-1}\right)\)
\(=\overline{......0}\)
\(\Rightarrow\)Chữ số tận cùng của \(A\)là \(0\)
Bài 3:
a)Ta có : \(C=2+2^2+2^3+...+2^{99}+2^{100}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)+...+2^{96}\left(2+2^2+2^3+2^4\right)\)
\(=31+2^4.31+...+2^{96}.31\)
\(=31\left(1+2^4+...+2^{96}\right)⋮31\)
\(\Rightarrow\)\(đpcm\)
b) Ta có : \(C=2+2^2+2^3+...+2^{99}+2^{100}\)
\(\Rightarrow2C=2^2+2^3+2^4+...+2^{100}+2^{101}\)
\(\Rightarrow2C-C=\left(2^2+2^3+2^4+...+2^{100}+2^{101}\right)-\left(2+2^2+2^3+...+2^{99}+2^{100}\right)\)
\(\Rightarrow C=2^{101}-2\)
Mà \(2^{2x}-2=C\)
\(\Rightarrow2^{2x}-2=2^{101}-2\)
\(\Rightarrow2^{2x}=2^{101}\)
\(\Rightarrow2x=101\)
\(\Rightarrow x=\frac{101}{2}\)
Vậy \(x=\frac{101}{2}\)
Bài 2:
Ta có : \(\overline{abcd}=1000a+100b+10c+d\)
\(=1000a+96b+8c+\left(d+2c+4b\right)\)
\(=8\left(125a+12b+c\right)+\left(d+2c+4b\right)\)
Vì \(\hept{\begin{cases}d+2c+4b⋮8\\8\left(125a+12b+c\right)⋮8\end{cases}}\)
\(\Rightarrow\overline{abcd}⋮8\)
\(\Rightarrowđpcm\)
a, C = 2 + 2 2 + 2 3 + ... + 2 99 + 2 100
2 C = 2 2 + 2 3 + 2 4 + ... + 2 100 + 2 101
2C - C = ( 2 2 + 2 3 + 2 4 + ... + 2 100 + 2 101 )
- ( 2 + 2 2 + 2 3 + ... + 2 99 + 2 100 )
C = 2 101 - 2
b , 2 2x - 1 - 2 = C
2 2x - 1 - 2 = 2 101 - 2
=> 2 2x - 1 = 2 101
=> 2x - 1 = 101
2x = 101 + 1
2x = 102
x = 51
c ) C = 2 + 2 2 + 2 3 + ... + 2 99 + 2 100
C = ( 2 + 2 2 + 2 3 + 2 4 + 2 5 ) + .... + ( 2 96 + 2 97 + 2 98 + 2 99 + 2 100 )
C = ( 2 + 2 2 + 2 3 + 2 4 + 2 5 ) + .... + ( 2 + 2 2 + 2 3 + 2 4 + 2 5 ) . 2 95
C = 31 + ..... + 31 . 2 95
C = 31 ( 1 + ... + 2 95 )
Vì 31 chia hết cho 31
=> C = 31 ( 1 + ... + 2 95 ) chia hết cho 31
Vậy C chia hết cho 31
a/ \(C=2+2^2+2^3+......+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+......+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+......+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+2^6.31+......+2^{96}.31\)
\(=31\left(2+2^6+.....+2^{96}\right)⋮31\left(đpcm\right)\)
b/ \(C=2+2^2+2^3+..........+2^{99}+2^{100}\)
\(\Leftrightarrow2C=2^2+2^3+.......+2^{100}+2^{101}\)
\(\Leftrightarrow2C-C=\left(2^2+2^3+......+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(\Leftrightarrow C=2^{101}-2\)
\(2^{2x-1}-2=C\)
\(\Leftrightarrow2^{2x-1}-2=2^{101}-2\)
\(\Leftrightarrow2^{2x}=2^{101}\)
\(\Leftrightarrow2x=101\)
\(\Leftrightarrow x=\dfrac{101}{2}\)
Vậy ..
a/C=2+2^2+2^3+.........+2^100
<=>2C-C=2(2+2^2+2^3+.....+2^100)-(2+2^2+2^3+.....+2^100)
C=2.2+2.2^2+2.2^3+....+2.2^100-(2+2^2+2^3+.....+2^100)
C=2^2+2^3+2^4+...+2^101-(2+2^2+2^3+.....+2^100)
Bạn loại các số giống nhau ta có:
C=2^101-2
b/C=2+2^2+2^3+.........+2^100
<=>C=(2+2^2+2^3+2^4+2^5)+......+(2^96+2^97+2^98+2^99+2^100)(nhóm 5 số lại nha)
=>C=62+2^6.(1+2+2^2+2^3+2^4)+....+2^96.(1+2+2^2+2^3+2^4)
<=>C=31.2+2^6.31+2^11.31+...+2^96.31
Đặt 31 làm thừa số chung.
C=31.(2+2^6+2^11+2^16+...+2^96) chia hết cho 31
Vậy C chia hết cho 31=>đpcm
c/22x-1-2=C
22x-1-2=2101-2
=>22x-1=2101
=>2x-1=101
2x =101-1
2x =100
x =100:2
x =50
a)ta có :C=2+2^2+2^3+...+2^99+2^100
C=2(1+2+2^2+2^3+2^4)+...+2^96(1+2+2^2+2^3+2^4)
C=2.31+...+2^96.31
C=31(2+...+2^96)
Vì 31 chia hết cho 31 nên suy ra 31(2+...+2^96) chia hết cho 31 hay C chia hết cho 31
Vậy C chia hết cho 31
b)Ta có:C=2+2^2+2^3+...+2^99+2^100
2C=2(2+2^2+2^3+...+2^99+2^100)
2C=2^2+2^3+2^4+...+2^100+2^101
2C-C=(2^2+2^3+2^4+...+2^100+2^101)-(2+2^2+2^3+...+2^99+2^100)
C=2^101-2
Mà 2^(2x-1)-2=C
Suy ra 2^(2x-1)=2^101
2x-1=101
2x=101+1
2x=102
x=102:2
x=51
Vậy x=51
a) C = 2 + 22 + 23 + ..... + 299 + 2100 ( có 100 số hạng )
C = ( 2 + 22 + 23 + 24 + 25 ) + ...... + ( 296 + 297 + 998 + 299 + 2100 )
C = 2 ( 1 + 2 + 22 + 23 + 24 ) + ..... + 296 ( 1 + 2 + 22 + 23 + 24 )
C = 2 . 31 + ..... + 296 . 31
C = 31 ( 2 + ..... + 296 )
vì 31 chia hết cho 31 => C chia hết cho 31
b) C = 2 + 22 + 23 + ..... + 299 + 2100
2C = 22 + 23 + 24 + ....... + 2100 + 2101
2C - C = 22 + 23 + 24 + ..... + 2100 + 2101 - ( 2 + 22 + 23 + ...... + 299 + 2100 )
C = 22 + 23 + 24 + ....... + 2100 + 2101 - 2 - 22 - 23 - ..... - 299 - 2100
C = 2101 - 2
=> 2x - 1 = 101
=> 2x = 101 + 1
=> 2x = 102
=> x = 102 : 2
=> x = 51
chúc bạn học giỏi ^.~
C=(2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)+......+(2^96+2^97+2^98+2^99+2^100)
C=2(1+2+2^2+2^3+2^4)+2^6(1+2+2^2+2^3+2^4)+......+2^96(1+2+2^2+2^3+2^4)
C=2.31+2^6.31+......+2^96.31
C=31(2+2^6+....+2^96) chia hết cho 31(đpcm)
chúc bạn học tốt
tick cho mk nha