Question

5²ⁿ −1 is divisible by 24 for all n ∈ N.

Answer 1

Let P(n) be the given statement.
Now,
$$\begin{gathered} P\left(n\right):5^{2n}-1\mathrm{is}\mathrm{divisible}\mathrm{by}24\mathrm{for}\mathrm{all}n\in N. \\ \mathrm{Step}1: \\ P\left(1\right)=5^2-1=25-1=24 \\ It\mathrm{is}\mathrm{divisible}\mathrm{by}24. \\ \mathrm{Thus},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)be\mathrm{true}. \\ \mathrm{Then},5^{2m}-1\mathrm{is}\mathrm{divisible}\mathrm{by}24. \\ Now,l\mathrm{et}{5}^{2m}-1=24\lambda ,\mathrm{where}\lambda \in N. \\ \mathrm{We}\mathrm{need}\mathrm{to}\mathrm{show}\mathrm{that}P(m+1)i\mathrm{s}\mathrm{true}\mathrm{whenever}P\left(m\right)\mathrm{is}\mathrm{true}. \\ Now, \\ P(m+1)=5^{2m+2}-1 \\ =5^{2m}{5}^2-1 \\ =25(24\lambda +1)-1 \\ =600\lambda +24 \\ =24(25\lambda +1) \\ Itis\mathrm{divisible}\mathrm{by}24. \\ \mathrm{Thus},P(m+1)istrue. \\ Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}i\mathrm{nduction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}n\in N. \end{gathered}$$