(n%5E2-3n-1)-2n%5E3%2B1-chia-h%E1%BA%BFt-cho-5.jpeg "(2n+1)(n^2-3n-1)-2n^3+1 Chia Hết Cho 5")
Proving (2n+1)(n^2-3n-1)-2n^3+1 is Divisible by 5
This article will demonstrate that the expression (2n+1)(n^2-3n-1)-2n^3+1 is always divisible by 5, for any integer value of 'n'.
Expanding the Expression
First, let's expand the expression:
(2n+1)(n^2-3n-1)-2n^3+1 = 2n^3 - 6n^2 - 2n + n^2 - 3n - 1 - 2n^3 + 1
Combining like terms, we get:
= -5n^2 - 5n
Factoring out a 5
Now, we can factor out a 5 from the expression:
= 5(-n^2 - n)
Conclusion
As we can see, the expression is a product of 5 and another integer (-n^2 - n). Since it is a product of 5, it is always divisible by 5 for any integer value of 'n'.
Therefore, we have proven that (2n+1)(n^2-3n-1)-2n^3+1 is divisible by 5 for all integer values of 'n'.