Use mathematical Induction to prove :...

MathematicsSum of Special SeriesJEE Advanced 1983Moderate

Use mathematical Induction to prove : If $n$ is any odd positive integer, then $n (n^{2} - 1)$ is divisible by 24.

Visualized Solution (English)

Defining the Odd Integer .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Let $n$ be an odd positive integer.
We can represent $n$ as $n = 2 m - 1$ , where $m \in {1, 2, 3, \dots}$ .
The expression becomes $P (m) = (2 m - 1) ((2 m - 1)^{2} - 1)$ .

Simplifying the Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (m)

Expand the squared term: $(2 m - 1)^{2} = 4 m^{2} - 4 m + 1$ .
Substitute back: $P (m) = (2 m - 1) (4 m^{2} - 4 m + 1 - 1)$ .
Simplify: $P (m) = (2 m - 1) (4 m^{2} - 4 m)$ .
Factor out $4 m$ : $P (m) = 4 m (m - 1) (2 m - 1)$ .

Base Case: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m = 1

For $m = 1$ :
Substitute $m = 1$ into $P (m) = 4 m (m - 1) (2 m - 1)$ .
$P (1) = 4 (1) (1 - 1) (2 (1) - 1) = 4 (1) (0) (1) = 0$ .
Since $0$ is divisible by $24$ , the base case is true.

Verification: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m = 2

For $m = 2$ :
$P (2) = 4 (2) (2 - 1) (2 (2) - 1)$ .
$P (2) = 4 (2) (1) (3) = 24$ .
Since $24$ is divisible by $24$ , the statement holds for $m = 2$ .

Inductive Hypothesis: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m = k

Assume the statement is true for $m = k$ .
$P (k) = 4 k (k - 1) (2 k - 1)$ is divisible by $24$ .
This means $P (k) = 24 λ$ for some integer $λ$ .

Target Expression: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m = k + 1

For $m = k + 1$ :
$P (k + 1) = 4 (k + 1) ((k + 1) - 1) (2 (k + 1) - 1)$ .
Simplify terms: $P (k + 1) = 4 (k + 1) (k) (2 k + 1)$ .

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (k + 1) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (k)

Expand $P (k + 1)$ : $4 k (k + 1) (2 k + 1) = 4 k (2 k^{2} + 3 k + 1) = 8 k^{3} + 12 k^{2} + 4 k$ .
Expand $P (k)$ : $4 k (k - 1) (2 k - 1) = 4 k (2 k^{2} - 3 k + 1) = 8 k^{3} - 12 k^{2} + 4 k$ .

The Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (k + 1) - P (k)

Calculate the difference: $P (k + 1) - P (k) = (8 k^{3} + 12 k^{2} + 4 k) - (8 k^{3} - 12 k^{2} + 4 k)$ .
$P (k + 1) - P (k) = 24 k^{2}$ .
Therefore, $P (k + 1) = P (k) + 24 k^{2}$ .

Final Proof of Divisibility

From the hypothesis, $P (k)$ is divisible by $24$ .
The term $24 k^{2}$ is clearly divisible by $24$ for any integer $k$ .
Since both terms are divisible by $24$ , their sum $P (k) + 24 k^{2}$ is also divisible by $24$ .
Thus, $P (k + 1)$ is divisible by $24$ .

Conclusion and Key Takeaway

Key Takeaway: For any odd integer $n$ , $n (n^{2} - 1)$ is a multiple of $24$ .
The Principle of Mathematical Induction confirms the statement for all $n \in {1, 3, 5, \dots}$ .
Next Challenge: Try proving divisibility by $48$ if $n$ is of the form $4 k \pm 1$ .

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