\(A=\left(2+2^2\right)+...+\left(2^{99}+2^{100}\right)\)
\(A=2\cdot\left(1+2\right)+...+2^{99}\cdot\left(1+2\right)\)
\(A=2\cdot3+...+2^{99}\cdot3\)
\(A=3\cdot\left(2+...+2^{99}\right)⋮3\left(đpcm\right)\)
2 ý kia tương tự
Giải:
Đặt S=(2+2^2+2^3+...+2^100)
=2.(1+2+2^2+2^3+2^4)+2^6.(1+2+2^2+2^3+2^4)+...+(1+2+2^2+2^3+2^4).296
=2.31+26.31+...+296.31
=31.(2+26+...+296)\(⋮\)31
Ta có :
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
=> \(A=(2+2^2)+(2^3+2^4)+...+(2^{99}+2^{100})\)
=> \(A=2(1+2)+2^3(1+2)+...+2^{99}(1+2)\)
=> \(A=2.3+2^3.3+...+2^{99}.3\)
=> \(A=(2+2^3+...+2^{99}).3\)chia hết cho 3 ( 1 )
Ta lại có :
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
=> \(A=2(1+2+2^2+2^3+...+2^{98}+2^{99})\)chia hết cho 2 ( 2 )
Từ ( 1 ) và ( 2 ) ta có :
A chia hết cho 2 . 3 hay A chia hết cho 6
Ta có :
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
=> \(A=\left(2+2^2+2^3+2^4+2^5\right)+....\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
=> \(A=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
=> \(A=2.31+...+2^{96}.31\)
=> \(A=\left(2+...+2^{96}\right)31\)chia hết cho 31
ta co A= [2+2^2] +[2^3+2^4] + .......... [2^99+2^100]
=3 .2 + 3.2^3 +............+3. 2^99
= 3 [2 + 2^3 +..........2^99] chia het 3 [dpcm]
ta co :A= [2+2^2] + [2^3 + 2^4] + ............[2^99+ 2^100]
= 6 . 1 + 6. 2^2 +..................6 .2^98
=6 . [1 + 2^2 +............2^98]chia het 6(dpcm)
ta co A= [2+2^2 +2^3 + 2^4 + 2^5] +................[2^96 + 2^ 97 + 2^ 98 +2^99 +2^100]
=31 . 2 + .....................+ 31 . 2^96 chia het 31(dpcm)
Ta có:
\(A=2^1+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(\Leftrightarrow A=2\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(\Leftrightarrow A=2\cdot3+...+2^{99}\cdot3\)
\(\Leftrightarrow A=3\left(2+...+2^{99}\right)⋮3\)
