1. Cho số nguyên x là 9 (Thỏa mãn x:7, dư 2); 2x+3(giả thuyết)

=> (2.9)+3 = 21 chia hết cho7 (chia hết cho viết bằng ki hiệu nha bạn)

2. 2^0+2^1+2^2+2^3+...+2^5n-3+2^5n-2+2^5-1

= (2^0+2^1+2^2+2^3+2^4)+...+(2^5n-5+2^5n-4+2^5n-3+2^5n-2+2^5n-1)

=(1+2+4+8+16)+...+(2^5n-5+2^5n-4+2^5n-3+2^5n-2+2^5n-1) chia hết cho 31

Đặt A = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ + ..... +2^{5n-6 +} 2^5n-5 + 2^5n-4 + 2^5n-3 + 2^5n-2 + 2^5n-1

=> A = ( 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ ) + ..... + ( 2^5n-6 + 2^5n-5 + 2^5n-4 + 2^5n-3 + 2^5n-2 + 2^{5n-1 )}

=> A = 2⁰ ( 1 + 2¹ + 2² + 2³ + 2⁴ ) + ..... + 2^5n-6 ( 1 + 2¹ + 2² + 2³ + 2⁴ )

=> A = 1.31 + 2⁵ .31 + ..... + 2^5n-6.31 

=> A = 31.( 1 + 2⁵ + ..... + 2^5n-6 )

Vì 31 ⋮ 31 => A ⋮ 31 ( đpcm )

\(\text{Đặt }A=\left(2^0+2^1+2^2+2^3+2^4\right)+...+\left(2^{5n-5}+2^{5n-4}+2^{5n-3}+2^{5n-2}+2^{5n-1}\right)\)

\(=\left(1+2 +4+8+16\right)+...+2^{5n-5}.\left(2^0+2^1+2^2+2^3+2^4\right)\)

\(=31+...+2^{5n-5}.31\)

\(=31.\left(1+...+2^{5n-5}\right)\text{chia hết cho 31}\left(đpcm\right)\)

Đặt A=\(2^0+2^1+2^2+....+2^{5n-3}+2^{5n-2}+2^{5n-1}\)

\(\Leftrightarrow A=\left(2^0+2^1+2^2+2^3+2^4+2^5\right)+...+\left(2^{5n+2}+2^{5n+1}+2^{5n}+2^{5n-1}+2^{5n-2}+2^{5n-3}\right)\)

\(\Leftrightarrow A=2^0\left(1+2+2^2+2^3+2^4\right)+....+2^{5n+2}\left(1+2+2^2+2^3+2^4\right)\)

\(\Leftrightarrow A=2^0\cdot31+2^5\cdot31+....+2^{5n+2}\cdot31\)

\(\Leftrightarrow A=31\cdot\left(2^0+2^5+...+2^{5n+2}\right)\)

\(\Rightarrow A⋮31\left(đpcm\right)\)

quynh oi dpcm la gi vay?

Là bạn muốn tớ thử đấy nhé!

Đặt S=1+2+2²+...+2^5n-3+2^5n-2+2^5n-1

=(1+2²+2⁴+2⁶+2⁸)+(2+2³+2⁵+2⁷+2⁹)+...+(2^5n-7+2^5n-5+2^5n-3+2^5n-1)

=1(1+2²+2⁴+2⁶+2⁸)+2(1+2²+2⁴+2⁶+2⁸)+...+2^5n-7(1+2²+2⁴+2⁶+2⁸)

=1.31+2.31+...+2^5n-7.31 chia hết cho 31

Vậy ...