Question

cho  A = 2 + 2^2 + 2^3 +..........+2^60 chứng tỏ rằng  A chia hết cho 3,7

Answer 1

TA có:VÌ 2= 2^1 

A=\(2^1+2^2+2^3+...+2^{60}\)

A= \(\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

A= \(2\left(1+2\right)+2^3\left(1+2\right)+...2^{59}\left(1+2\right)\)

A= \(3.\left(2+2^3+...+2^{60}\right)\)chia hết cho 3

=) A chia hết cho3( đpcm)

Ta lại có:

A= \(2^1+2^2+2^3+...+2^{60}\)

A= \(\left(2^1+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

A=\(2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

A= \(7.\left(2+...+2^{58}\right)\)chia hết cho 7

=) A chia hết cho 7( đpcm)

Answer 2

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