số 17

17

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A=1+3+3^2+3^3+...+3^20

3A=3+3^2+3^3+...+3^21

3A - A=(3+3^2+3^3+...+3^21) - 1+3+3^2+3^3+...+3^20

2A=3^21 - 1

ok

1

Ta có: 5+ 5² + 5³ + ... + 5¹²
= 1.(5+5²)+5²(5+5²)+ ... +5¹⁰(5+5²)

=1.30 + 5².30+ ... +5¹⁰.30

=30.(1 + 5² + ... + 5¹⁰) ⋮ 30

Vậy 5 + 5² + 5³ + ... + 5¹² ⋮ 30

5+5²+5³+...+5¹²

=(5+5²)+(5³+5⁴)+...+(5¹¹+5¹²)

=(5.1+5².1)+(5².5+5².5²)+...+(5¹⁰.5+5¹⁰.5²)

=1.(5+5²)+5².(5+5²)+...+5¹⁰.(5+5²)

=1.30+5².30+...+5¹⁰.30

=30.(1+5²+...+5¹⁰):30

Vậy (5+5²+5³+...+5¹²) chia hết cho 30

`(-x+5).(6-x)=0`

`=>`\(\left[{}\begin{matrix}-x+5=0\\6-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

`A=(3+3^{2})+(3^{3}+3^{4})+(3^{5}+3^{6})+...+(3^{59}+3^{60})`

`=3.(1+3)+3^{3}.(1+3)+3^{5}.(1+3)+...+3^{59}.(1+3)`

`=3.4+3^{3}.4+3^{5}.4+...+3^{59}.4`

`=4.(3+3^{3}+3^{5}+...+3^{59})\vdots 4`

Ta có A = 3 + 3² + 3³ + ... + 3⁶⁰

= ( 3 + 3² ) + ( 3³ + 3⁴ ) + ... + ( 3⁵⁹ + 3⁶⁰ )

= 3( 1 + 3 ) + 3³( 1 + 3 ) + ... + 3⁵⁹( 1 + 3 )

= 3 . 4 + 3³ . 4 + ... + 3⁵⁹ . 4

= 4( 3 + 3³ + ... + 3⁵⁹ ) ⋮ 4 vì 4 ⋮ 4

Vậy A ⋮ 4

Ta có A = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰

= ( 2 + 2² ) + ( 2³ + 2⁴ ) + ... + ( 2⁹⁹ + 2¹⁰⁰ )

= 2( 1 + 2 ) + 2³( 1 + 2 ) + .... + 2⁹⁹( 1 + 2 )

= 2 . 3 + 2³ . 3 + ... + 2⁹⁹ . 3

= 3( 2 + 2³ + ... + 2⁹⁹ ) ⋮ 3 vì 3 ⋮ 3

Vậy A ⋮ 3

Ta có A = 3 + 3² + 3³ + ... + 3¹²⁰

= ( 3 + 3² + 3³ ) + ( 3⁴ + 3⁵ + 3⁶ ) + ... + ( 3¹¹⁸ + 3¹¹⁹ + 3¹²⁰ )

= 3( 1 + 3 + 3² ) + 3⁴( 1 + 3 + 3² ) + ... + 3¹¹⁸( 1 + 3 + 3² ) 

= 3 . 13 + 3⁴ . 13 + ... + 3¹¹⁸ . 13

= 13( 3 + 3⁴ + ... + 3¹¹⁸ ) ⋮ 13 vì 13 ⋮ 13

Vậy A ⋮ 13

    12.75 + 12.17 + 12.18

= 12. (75 + 17 + 18)

=12. 110

= 1320