Mathematical Induction

 

Prove by mathematical induction \[ 5^{2n} - 1 \] is divisible by 24 for all natural numbers \( n \gt 0 \).

Prove the statement is true for a particular n. Here n = 1 is required.

To show this is true for n=1, observe that the expression = 24. Now 24 is divisible by 24, so the statement is true for n = 1.

Assume the statement is true for n = k and show it is true for n = k + 1.

To show the statement is true for n = k+1, \[ \begin{align} 5^{2(k+1)} - 1 &= 5^{2k+2} - 1 \\ &= 5^{2k} \times 5^2 - 1 \\ \end{align} \ \]

Use the inductive hypothesis to simplify the right hand side.

By the inductive hypothesis, \[ \begin{align} 5^{2k} \times 5^2 - 1 &= (24m + 1) \times 25 - 1 \quad \text{for some integer m}\\ &= 24 \times 25 m + 25 - 1 \\ &= 24(25m + 1) \end{align} \ \] Which is divisible by 24. So the statement is true for n = k+1.

By the principle of mathematical induction,

The statement is true for \( n \gt 0 \).

You will gain most benefit if you work through each problem first and then come back and check your work against the solution.

 

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Now You Try:

Prove by mathematical induction \[ n^2 + n \] is even for all natural numbers \( n \gt 0 \).
There is an easier way to prove this. What is it?