Question

2.7ⁿ + 3.5ⁿ − 5 is divisible by 24 for all n ∈ N.

Answer 1

Let P(n) be the given statement.
Now,
$$\begin{gathered} P\left(n\right):2.7^n+3.5^n-5\mathrm{is}\mathrm{divisible}\mathrm{by}24. \\ \mathrm{Step}1: \\ P\left(1\right):2.7^1+3.5^1-5=24 \\ It\mathrm{is}\mathrm{divisible}\mathrm{by}24. \\ \mathrm{Thus},P\left(1\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)\mathrm{be}\mathrm{true}. \\ Then, \\ 2.7^m+3.5^m-5\mathrm{is}\mathrm{divisible}\mathrm{by}24. \\ \mathrm{Suppose}: \\ 2.7^m+3.5^m-5=24\lambda ...\left(1\right) \\ \mathrm{We}\mathrm{need}\mathrm{to}\mathrm{show}\mathrm{that}P\left(m+1\right)istrue\mathrm{whenever}P\left(m\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Now}, \\ P\left(m+1\right)=2.7^{m+1}+3.5^{m+1}-5 \\ =2.7^{m+1}+(24\lambda +5-2.7^m)5-5 \\ =\left(2.7^{m+1}+120\lambda +25-10.7^m-5\right) \\ =2.7^m.7-10.7^m+120\lambda +24-4 \\ =7^m\left(14-10\right)+120\lambda +24-4 \\ =7^m.4+120\lambda +24-4 \\ =4\left(7^m-1\right)+24\left(5\lambda +1\right) \\ =4\times 6\mu +24(5\lambda +1)\left[\mathrm{Since}{7}^m-1\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}6\mathrm{for}\mathrm{all}n\in N,7^m-1=\mu .\right] \\ =24(\mu +5\lambda +1) \\ It\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}24. \\ \mathrm{Thus},P(m+1)\mathrm{is}\mathrm{true}. \\ Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}i\mathrm{nduction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}n\in N. \end{gathered}$$