Final answer:
By expanding the cubes and algebraically manipulating the terms, it's proved that for any natural number n≥2, n^3 + (n+1)^3 is greater than (n+2)^3.
Step-by-step explanation:
To prove that n3 + (n+1)3 > (n+2)3 for all n ∈ N with n ≥ 2, we will apply some basic algebraic manipulations and the properties of cubing of exponentials.
First, let's expand the cubes using the binomial theorem:
- n3 + (n+1)3 = n3 + n3 + 3n2 + 3n + 1
- (n+2)3 = n3 + 3n2 + 6n + 8
Now, subtract (n+2)3 from n3 + (n+1)3:
- n3 + n3 + 3n2 + 3n + 1 - (n3 + 3n2 + 6n + 8) = 2n3 - 3n2 - 3n - 7
To prove the inequality, we can show that for n≥2, the term 2n3 is larger than the combined terms -3n2 - 3n - 7:
And since n is a natural number starting from 2, this is clearly true as the cube term grows much faster than the quadratic and linear terms combined with the constant.
Therefore, n3 + (n+1)3 indeed is greater than (n+2)3 for all n∈N with n≥2.