Prove that for any positive integer $n$, $n^3 - n$ is divisible by $6$.
Step-by-Step Solution
Key Concept: $n^3 - n = n(n^2 - 1) = (n - 1)n(n + 1)$, which is the product of three consecutive integers.
$n^3 - n = n(n-1)(n+1) = (n-1)n(n+1)$. This is the product of 3 consecutive integers. [1.0 Mark]
Among any 3 consecutive integers, at least one is divisible by $2$ and exactly one is divisible by $3$. [1.0 Mark]
Since $2$ and $3$ are coprime, the product is divisible by $2 \times 3 = 6$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Factoring $n^3 - n = (n-1)n(n+1)$: 1.0 Mark
Arguing product of 3 consecutive integers has factors 2 and 3: 1.0 Mark
Concluding divisibility by 6: 1.0 Mark
Correct Answer: