1. A = 2 + 2² + 2³ + 2⁴ + ... + 2⁶⁰

A = ( 2 + 2² + 2³ ) + ( 2⁴ + 2⁵ + 2⁶ ) + ... + ( 2⁵⁸ + 2⁵⁹ + 2⁶⁰ )

A = 2 ( 1 + 2 + 2² ) + 2⁴ ( 1 + 2 + 2² ) + ... + 2⁵⁸ ( 1 + 2 + 2² )

A = 2 . 7 + 2⁴ . 7 + ... + 2⁵⁸ . 7

A = ( 2 + 2⁴ + ... + 2⁵⁸ ) . 7 => A \(⋮\)7

Vậy ...

2.Ta có : \(n+4⋮n+1\)

Mà : \(n+1⋮n+1\)

\(\Rightarrow\left(n+4\right)-\left(n+1\right)⋮n+1\Rightarrow n+4-n-1⋮n+1\)

\(\Rightarrow3⋮n+1\Rightarrow n+1\in\left\{1;3\right\}\)

\(\Rightarrow n\in\left\{0;2\right\}\)

3. Đặt B = 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷

B = ( 1 + 2 ) + ( 2² + 2³ ) + ( 2⁴ + 2⁵ ) + ( 2⁶ + 2⁷ )

B = ( 1 + 2 ) + 2² ( 1 + 2 ) + 2⁴ ( 1 + 2 ) + 2⁶ ( 1 + 2 )

B = 1 . 3 + 2² . 3 + 2⁴ . 3 + 2⁶ . 3

B = ( 1 + 2² + 2⁴ + 2⁶ ) . 3 \(\Rightarrow\) B \(⋮\)3

Vậy ...

1) Ta có: \(2⋮n-3\)

\(\Leftrightarrow n-3\inƯ\left(2\right)\)

\(\Leftrightarrow n-3\in\left\{1;-1;2;-2\right\}\)

hay \(n\in\left\{4;2;5;1\right\}\)

Vậy: \(n\in\left\{4;2;5;1\right\}\)

2) Ta có: \(n+2⋮n-3\)

\(\Leftrightarrow n-3+5⋮n-3\)

mà \(n-3⋮n-3\)

nên \(5⋮n-3\)

\(\Leftrightarrow n-3\inƯ\left(5\right)\)

\(\Leftrightarrow n-3\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{4;2;8;-2\right\}\)

Vậy: \(n\in\left\{4;2;8;-2\right\}\)

ko biết

1. \(A=\frac{1}{2}-\frac{2}{5}+\frac{1}{3}+\frac{5}{7}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)

\(=\frac{1}{2}-\frac{2}{5}+\frac{1}{3}+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)

\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)-\left(\frac{2}{5}-\frac{5}{7}+\frac{4}{35}\right)+\frac{1}{41}\)

\(=\left(\frac{5}{6}+\frac{1}{6}\right)-\left(\frac{-11}{35}+\frac{4}{35}\right)+\frac{1}{41}\)\(=1-\frac{-7}{35}+\frac{1}{41}=1+\frac{1}{5}+\frac{1}{41}=\frac{251}{205}\)

2. a) \(1+4+4^2+4^3+......+4^{99}=\left(1+4\right)+\left(4^2+4^3\right)+.......+\left(4^{98}+4^{99}\right)\)

\(=\left(1+4\right)+4^2\left(1+4\right)+.........+4^{98}\left(1+4\right)\)

\(=5+4^2.5+........+4^{98}.5=5\left(1+4^2+.....+4^{98}\right)⋮5\)( đpcm )

b) \(3^{n+2}-2^{n+2}+3^n-2^n=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)

\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)=3^n\left(9+1\right)-2^n\left(4+1\right)\)

\(=3^n.10-2^n.5=3^n.10-2^{n-1+1}.5=3^n.10-2^{n-1}.2.5\)

\(=3^n.10-2^{n-1}.10=10\left(3^n-2^{n-1}\right)⋮10\)( đpcm )

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