Final answer:
The expression ⅓ n³ + ½ n² + ⅙ n is an integer for any natural number n because it can be factored to show that it consists of the product of three consecutive integers, which together with a separate term divisible by 6, guarantees the result is an integer.
Step-by-step explanation:
To prove that the expression ⅓ n³ + ½ n² + ⅙ n is an integer for any n ∈ N, we need to factor out n from the expression, getting:
n(⅓ n² + ½ n + ⅙)
Now, notice that the expression inside the brackets is the same as:
⅓ n² + ⅓ n + ⅓ with another ⅓ n² + ⅚
Or, more clearly:
⅓ n(n - 1)(n + 1) + ⅚ n)
This breaks down to the sum of two parts: ⅓ n(n - 1)(n + 1) is always an integer because it is the product of three consecutive numbers, one of which must be a multiple of 3, ensuring that the fraction is eliminated. The second part, ⅚ n, is also an integer since any natural number n divided by 6 yields an integer or zero. Combining both parts, we establish that the entire expression is an integer.
Therefore, we have proved that given expression is always an integer when n is a natural number.