Question

Cho Y = 1+3+3^2+3^3+...+3^98.   Chứng tỏ rằng Y chia hết cho 13  

giải giúp mình với , mình cần gấp .

giải logic ra giúp mình nhé 

cảm ơn trước 

Answer 1

\(Y=1+3+3^2+3^3+.......+3^{98}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+.........+\left(3^{96}+3^{97}+3^{98}\right)\)

\(=\left(1+3+3^2\right)+3^3.\left(1+3+3^2\right)+......+3^{96}.\left(1+3+3^2\right)\)

\(=\left(1+3+9\right)+3^3.\left(1+3+9\right)+.........+3^{96}.\left(1+3+9\right)\)

\(=13+3^3.13+.......+3^{96}.13\)

\(=13.\left(1+3^3+.......+3^{96}\right)⋮13\)( đpcm )

Answer 2

Y = 1 + 3 + 3² + 3³ + ... + 3⁹⁸

= ( 1 + 3 + 3² ) + ( 3³ + 3⁴ + 3⁵ ) + ... + ( 3⁹⁶ + 3⁹⁷ + 3⁹⁸ )

= 13 + 3³( 1 + 3 + 3² ) + ... + 3⁹⁶( 1 + 3 + 3² )

= 13 + 3³.13 + ... + 3⁹⁶.13

= 13( 1 + 3³ + ... + 3⁹⁶ ) chia hết cho 13 ( đpcm )

Answer 3

Ta có:

\(Y=1+3+3^2+3^3+...+3^{98}\)(Có 99 số hạng)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\)(Có 33 nhóm)

\(=\left(1+3+3^2\right)+\left(1+3+3^2\right).3^3+...+3^{96}.\left(1+3+3^2\right)\)

\(=13+13.3^3+...+13.3^{96}\)

\(=13.\left(1+3^3+...+3^{96}\right)\)

Vì\(13⋮13\)

\(1+3^3+...+3^{96}\inℤ\)

Suy ra:\(13.\left(1+3^3+...+3^{96}\right)⋮13\)

Hay\(Y⋮13\left(đpcm\right)\)

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