7.1 Q-27 | Mathematical Induction And Binomial Theorem | Mathematics 11

Exercise 7.1 — Question 27

Problem Statement

Using the Binomial Theorem, prove that $n^{3} - n$ is divisible by $6$ for every positive integer $n$ .

Key Concept

To test divisibility using the Binomial Theorem, we rewrite the base in the form $(1 + k)^{n}$ or factor the expression so that multiples of the divisor become visible.

SLO Covered: Use Binomial Theorem to test divisibility of a number (M-11-A-40).

Solution

Step 1: Factor the expression.

$n^{3} - n = n (n^{2} - 1) = n (n - 1) (n + 1) = (n - 1) \cdot n \cdot (n + 1)$

This is the product of three consecutive integers.

Step 2: Apply divisibility reasoning.

Among any three consecutive integers $(n - 1), n, (n + 1)$ :

At least one is divisible by $2$ (even number).
Exactly one is divisible by $3$ .

Therefore, their product is always divisible by $2 \times 3 = 6$ .

$∴ 6 ∣ (n^{3} - n) for all n \in Z^{+}$

Alternative Approach Using Binomial Theorem

Write $n = (1 + (n - 1))$ . Then:

$n^{3} = (1 + (n - 1))^{3} = \sum_{r = 0}^{3} (r 3) (n - 1)^{r}$

$= 1 + 3 (n - 1) + 3 (n - 1)^{2} + (n - 1)^{3}$

So: $n^{3} - n = 1 + 3 (n - 1) + 3 (n - 1)^{2} + (n - 1)^{3} - n$

Simplifying: $= 3 (n - 1) + 3 (n - 1)^{2} + (n - 1)^{3} - (n - 1)$ $= (n - 1) [3 + 3 (n - 1) + (n - 1)^{2} - 1]$ $= (n - 1) [(n - 1)^{2} + 3 (n - 1) + 2]$ $= (n - 1) (n - 1 + 1) (n - 1 + 2)$ $= (n - 1) \cdot n \cdot (n + 1)$

This confirms the factored form, and the divisibility by 6 follows as shown above.

Summary

Expression	Divisible by	Reason
$n^{3} - n = (n - 1) n (n + 1)$	$6$	Product of 3 consecutive integers

Tip: Whenever a question asks to prove divisibility by $6$ , look for a product of three consecutive integers or show divisibility by both $2$ and $3$ separately.

This is the product of three consecutive integers.

Step 2: Apply divisibility reasoning.

Among any three consecutive integers $(n - 1), n, (n + 1)$ :

At least one is divisible by $2$ (even number).
Exactly one is divisible by $3$ .

Therefore, their product is always divisible by $2 \times 3 = 6$ .

$∴ 6 ∣ (n^{3} - n) for all n \in Z^{+}$

Alternative Approach Using Binomial Theorem

Write $n = (1 + (n - 1))$ . Then:

$n^{3} = (1 + (n - 1))^{3} = \sum_{r = 0}^{3} (r 3) (n - 1)^{r}$

$= 1 + 3 (n - 1) + 3 (n - 1)^{2} + (n - 1)^{3}$

So: $n^{3} - n = 1 + 3 (n - 1) + 3 (n - 1)^{2} + (n - 1)^{3} - n$

This confirms the factored form, and the divisibility by 6 follows as shown above.

Summary

Expression	Divisible by	Reason
$n^{3} - n = (n - 1) n (n + 1)$	$6$	Product of 3 consecutive integers

Tip: Whenever a question asks to prove divisibility by $6$ , look for a product of three consecutive integers or show divisibility by both $2$ and $3$ separately.