khó nhỉ

Bài 1.

2n²( n + 1 ) - 2n( n² + n - 3 )

= 2n³ + 2n² - 2n³ - 2nⁿ + 6n

= 6n \(⋮6\forall n\inℤ\)( đpcm )

Bài 2.

P = ( m² - 2m + 4 )( m + 2 ) - m³ + ( m + 3 )( m - 3 ) - m² - 18

P = m³ + 8 - m³ + m² - 9 - m² - 18

P = 8 - 9 - 18 = -19

=> P không phụ thuộc vào biến M ( đpcm )

\(1,\)

\(a,\) Với \(n=1\Leftrightarrow5+2\cdot1+1=8⋮8\left(đúng\right)\)

Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow5^k+2\cdot3^{k-1}+1⋮8\)

Với \(n=k+1\)

\(5^n+2\cdot3^{n-1}+1=5^{k+1}+2\cdot3^k+1\\ =5^k\cdot5+2\cdot3^k+1\\ =5^k\cdot2+2\cdot3^k+5^k\cdot3+1\\ =2\left(5^k+3^k\right)+5^k+2\cdot5^{k-1}+1+2\cdot3^{k-1}-2\cdot3^{k-1}\\ =2\left(5^k+3^k\right)+\left(5^k+2\cdot3^{k-1}+1\right)-2\left(3^{k-1}+5^{k-1}\right)\)

Vì \(5^k+3^k⋮\left(5+3\right)=8;5^{k-1}+3^{k-1}⋮\left(5+3\right)=8;5^k+2\cdot3^{k-1}+1⋮8\) nên \(5^{k+1}+2\cdot3^k+1⋮8\)

Theo pp quy nạp ta được đpcm

\(b,\) Với \(n=1\Leftrightarrow3^3+4^3=91⋮13\left(đúng\right)\)

Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow3^{k+2}+4^{2k+1}⋮13\)

Với \(n=k+1\)

\(3^{n+2}+4^{2n+1}=3^{k+3}+4^{2k+3}\\ =3^{k+2}\cdot3+16\cdot4^{2k+1}\\ =3^{k+2}\cdot3+3\cdot4^{2k+1}+13\cdot4^{2k+1}\\ =3\left(3^{k+2}+4^{2k+1}\right)+13\cdot4^{2k+1}\)

Vì \(3^{k+2}+4^{2k+1}⋮13;13\cdot4^{2k+1}⋮13\) nên \(3^{k+3}+4^{2k+3}⋮13\)

Theo pp quy nạp ta được đpcm

\(1,\)

\(c,C=6^{2n}+3^{n+2}+3^n\\ C=36^n+3^n\cdot9+3^n\\ C=\left(36^n-3^n\right)+\left(3^n\cdot9+2\cdot3^n\right)\\ C=\left(36^n-3^n\right)+3^n\cdot11\)

Vì \(36^n-3^n⋮\left(36-3\right)=33⋮11;3^n\cdot11⋮11\) nên \(C⋮11\)

\(d,D=1^n+2^n+5^n+8^n\)

Vì \(1^n+2^n+5^n⋮\left(1+2+5\right)=8;8^n⋮8\) nên \(D⋮8\)

\(2,\)

Ta thấy:\(1+2+...+2002=\left(2002+1\right)\left(2002-1+1\right):2=2003\cdot2002:2⋮11\left(2002⋮11\right)\)

Do đó \(1^{2002}+2^{2002}+...+2002^{2002}⋮1+2+...+2002⋮11\)

Ta có:3ⁿ⁺²+3ⁿ⁺¹+2ⁿ⁺³+2ⁿ⁺²=3ⁿ.(3³+3¹)+2ⁿ.(2³+2²)

3ⁿ⁺²+3ⁿ⁺¹+2ⁿ⁺³+2ⁿ⁺²=3ⁿ.30+2ⁿ.12=3ⁿ.6.5+2ⁿ.6.2

3ⁿ⁺²+3ⁿ⁺¹+2ⁿ⁺³+2ⁿ⁺²=6.(3ⁿ.5+2ⁿ.2)

⇒đpcm

Sửa đề \(3^{n+2}\rightarrow3^{n+3}\)

Giải:

Gọi \(M=3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)

Ta có:

\(M=3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\) 

\(M=3^n.3^3+3^n.3^1+2^n.2^3+2^n.2^2\) 

\(M=3^n.\left(27+3\right)+2^n.\left(8+4\right)\) 

\(M=3^n.30+2^n.12\) 

Vì 30 ⋮ 6 và 12 ⋮ 6

Nên \(3^n.30+2^n.12⋮6\) 

Vậy \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\left(đpcm\right)\)

A=3^n+3+2^n+3+3^n+1+2^n+2

A=(3^n+3+3^n+1)+(2^n+3+2^n+2)

A=3^n(3^3+3)+2^n(2^3+2^2)

=3^n.30+2^n.12

=6(3^n.5+2^n.2) chia hết cho 6

=>A chia hết cho 6

(Công nhận Nhi giỏi thật mới thi hôm qua mà tối hôm kia đã hỏi)

\(3^{n+2}-2^{n+2}+3^n-2^n\)

\(=\left(3^n.3^2+3^n\right)-\left(2^n.2^2+2^n\right)\)

\(=\left(3^n.10\right)-\left(2^n.5\right)=\left(3^n.10\right)-\left(2^{n-1}.10\right)\)

\(=\left(3^n-2^{n-1}\right).10⋮10\)

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