Hint : Since you probably missed some degrees in the expression ( 8 n − 14 n + 27 ) (8n-14n+27) ( 8 n − 14 n + 27 ) , I can only complete part of the task.
Using mathematical induction, we prove that ( n 3 + 11 n ) (n^3+11n) ( n 3 + 11 n ) is divisible by 6 6 6 for n ∈ Z + n\in\mathbb{Z}^+ n ∈ Z + .
1 STEP: Induction basis
n = 1 → 1 3 + 11 ⋅ 1 = 12 = 2 ⋅ 6 ⋮ 6 n=1\to 1^3+11\cdot1=12=2\cdot6\,\,\,\vdots\,\,\,6 n = 1 → 1 3 + 11 ⋅ 1 = 12 = 2 ⋅ 6 ⋮ 6
2 STEP : Inductive guess
For all k ≤ n ( k 3 + 11 k ) ⋮ 6 \text{For all}\,\,\,k\le n\,\,\,(k^3+11k)\,\,\,\vdots\,\,\,6 For all k ≤ n ( k 3 + 11 k ) ⋮ 6
3 STEP : Inductive transition, it is necessary to prove that
For k = n + 1 → ( ( n + 1 ) 3 + 11 ( n + 1 ) ) ⋮ 6 \text{For}\,\,\,k=n+1\to \left((n+1)^3+11(n+1)\right)\,\,\,\vdots\,\,\,6 For k = n + 1 → ( ( n + 1 ) 3 + 11 ( n + 1 ) ) ⋮ 6
( n + 1 ) 3 + 11 ( n + 1 ) = n 3 + 3 n 2 + 3 n + 1 + 11 n + 11 = = ( n 3 + 11 n ) + ( 3 n 2 + 3 n ) + 12 = = ( n 3 + 11 n ) ⏟ divided into 6 hypothesis + 3 n ( n + 1 ) + 6 ⋅ 2 (n+1)^3+11(n+1)=n^3+3n^2+3n+1+11n+11=\\[0.3cm]
=(n^3+11n)+(3n^2+3n)+12=\\[0.3cm]
=\underbrace{(n^3+11n)}_{\text{divided into 6 hypothesis}}+3n(n+1)+6\cdot 2 ( n + 1 ) 3 + 11 ( n + 1 ) = n 3 + 3 n 2 + 3 n + 1 + 11 n + 11 = = ( n 3 + 11 n ) + ( 3 n 2 + 3 n ) + 12 = = divided into 6 hypothesis ( n 3 + 11 n ) + 3 n ( n + 1 ) + 6 ⋅ 2
It remains to understand why 3 n ( n + 1 ) ⋮ 6 3n(n+1)\,\,\,\vdots\,\,\,6 3 n ( n + 1 ) ⋮ 6 .
1 case : n = 2 k n=2k n = 2 k
3 n ( n + 1 ) = 3 ⋅ ( 2 k ) ⋅ ( 2 k + 1 ) = 6 ⋅ ( k ( 2 k + 1 ) ) ⋮ 6 3n(n+1)=3\cdot(2k)\cdot(2k+1)=6\cdot\left(k(2k+1)\right)\,\,\,\vdots\,\,\,6 3 n ( n + 1 ) = 3 ⋅ ( 2 k ) ⋅ ( 2 k + 1 ) = 6 ⋅ ( k ( 2 k + 1 ) ) ⋮ 6
2 case : n = 2 k + 1 n=2k+1 n = 2 k + 1
3 n ( n + 1 ) = 3 ⋅ ( 2 k + 1 ) ⋅ ( 2 k + 1 + 1 ) = 6 ⋅ ( ( k + 1 ) ( 2 k + 1 ) ) ⋮ 6 3n(n+1)=3\cdot(2k+1)\cdot(2k+1+1)=6\cdot\left((k+1)(2k+1)\right)\,\,\vdots\,\,6 3 n ( n + 1 ) = 3 ⋅ ( 2 k + 1 ) ⋅ ( 2 k + 1 + 1 ) = 6 ⋅ ( ( k + 1 ) ( 2 k + 1 ) ) ⋮ 6
Conclusion,
( n + 1 ) 3 + 11 ( n + 1 ) = ( n 3 + 11 n ) + 3 n ( n + 1 ) + 6 ⋅ 2 ⋮ 6 (n+1)^3+11(n+1)=(n^3+11n)+3n(n+1)+6\cdot2\,\,\vdots\,\,6 ( n + 1 ) 3 + 11 ( n + 1 ) = ( n 3 + 11 n ) + 3 n ( n + 1 ) + 6 ⋅ 2 ⋮ 6 Q.E.D.
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