Với n = 0

\(\Rightarrow3.5^{2.0+1}+2^{3.0+1}=3.5+2=15+2=17⋮17\Rightarrow\)đúng với n = 0

Giả sử \(3.5^{2n+1}+2^{3n+1}\) đúng với n = k \(\in\) N*

\(\Rightarrow3.5^{2k+1}+2^{3k+1}⋮17\)

C/m : \(3.5^{2n+1}+2^{3n+1}\) đúng với n = k + 1 ( k \(\in\) N* )

Ta có :

\(3.5^{2n+1}+2^{3n+1}=3.5^{2\left(k+1\right)+1}+2^{3\left(k+1\right)+1}\)

\(=3.25.5^{2k+1}+8.3^{3k+1}=3.25.5^{2k+1}+25.2^{3k+1}-17.2^{3k+1}\)

\(=25\left(3.5^{2k+1}+2^{3k+1}\right)-17.2^{3k+1}\)

Vì : \(17.2^{3k+1}⋮17\) ; \(3.5^{2k+1}+2^{3k+1}⋮17\) theo phương pháp quy nạp

\(\Rightarrow3.5^{2\left(k+1\right)+1}+2^{3\left(k+1\right)+1}⋮17\)

Vậy ...

Giải:

Ta có:

\(3^{4n+2}=9.9^{2n}=\) \(9.\left(17-8\right)^{2n}=17k+9.64^n\)

\(2.4^{3n+1}=8.64^n\)

\(\Rightarrow3^{4n+2}+2.4^{3n+1}=17k+17.64^n\)

\(=17\left(k+64^n\right)⋮17\forall x\in N\) (Đpcm)

a, Ta có:

\(3^{2n+1}+2^{n+2}=9^n.3+2^n.4\)

\(=9^n.3-2^n.3+2^n.7=3\left(9^n-2^n\right)+2^n.7\)

Ta lại có:

\(9^n-2^n⋮9-2=7;2n.7⋮7\)

\(\Rightarrow3^{2n+1}+2^{n+2}⋮7\left(dpcm\right)\)

a) n(n + 5) - (n - 3)(n + 2) = n² + 5n - n² - 2n + 3n + 6 = 6n + 6 = 6(n + 1) \(⋮\)6 \(\forall\)x \(\in\)Z

b) (n² + 3n - 1)(n + 2) - n³  + 2 = n³ + 2n² + 3n² + 6n - n - 2 - n³ + 2 = 5n² + 5n = 5n(n + 1) \(⋮\)5 \(\forall\)x \(\in\)Z

c) (6n + 1)(n + 5) - (3n + 5)(2n - 1) = 6n² + 30n + n + 5 - 6n² + 3n - 10n + 5 = 24n + 10 = 2(12n + 5) \(⋮\)2 \(\forall\)x \(\in\)Z

d) (2n - 1)(2n + 1) - (4n - 3)(n - 2) - 4 = 4n² - 1 - 4n² + 8n + 3n - 6 - 4 = 11n - 11 = 11(n - 1) \(⋮\)11 \(\forall\)x \(\in\)Z

Trả lời ngắn tí như ri này:

Ta có:\(3.25^n.5\) =\(15.25^n\) \(\equiv15.8^n\left(mod17\right)\) .

\(2^{3n+1}=8^n.2\left(mod17\right)\) .

\(\Rightarrow3.5^{2n+1}+2^{3n+1}\equiv15.8^n+2.8^n\left(mod17\right)\) .

\(=17.8^n\) chia hết cho 17 \(\forall\) so nguyên n.

\(3\cdot5^{2n+1}+2^{3n+1}=3\cdot5^{2n}\cdot5+2^{3n}\cdot2=15\cdot25^n+8^n\cdot2\)

\(=\left(17-2\right)\cdot25^n+8^n\cdot2=17\cdot25^n-2\cdot25^n+8^n\cdot2=17\cdot25^n-2\left(25^n-8^n\right)\)

\(=17\cdot25^n-2\left(25-8\right)\left(25^{n-1}+25^{n-2}\cdot8+25^{n-3}\cdot8^2+...+8^{n-1}\right)\)

\(=17\cdot25^n-34\left(25^{n-1}+25^{n-2}\cdot8+25^{n-3}\cdot8^2+...+8^{n-1}\right)\)

vì 17 chia hết cho 17 nên 17*25^n chia hết cho 17(1)

vì 34 chia hts cho 17 nên 34(25^n-1+25^n-2*8+25^n-3*8^2+...+8^n-1) chia hết cho 17

\(\Rightarrow17\cdot25^n-34\left(25^{n-1}+25^{n-2}\cdot8+25^{n-3}\cdot8^2+...+8^{n-1}\right)\)chia hết cho 17

\(\Rightarrow3\cdot5^{2n+1}+2^{3n+1}\)chia hết cho 17 (đpcm)

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

\(A=2n^3+6n^2+4n-n^2-3n-2-2n^3+4n+2\)

\(A=5n^2+5n\)

\(A=5n\left(n+1\right)\)

\(\text{Vì 5⋮5 nên 5n(n+1)⋮5}\)(1)

\(\text{Vì n;n+1 là hai số tự nhiên liên tiếp nên n(n+1)⋮2}\)

\(\Rightarrow5n\left(n+1\right)⋮2\)(2)

\(\text{Từ (1) và (2)}\Rightarrow5n\left(n+1\right)⋮10\text{ vì (2,5)=1}\)

\(\text{Vậy A⋮10}\)