Question

11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by 133 for all n ∈ N.

Answer 1

Let P(n) be the given statement.
Now,
$$\begin{gathered} P\left(n\right):{11}^{n+2}+{12}^{2n+1}\mathrm{is}\mathrm{divisible}\mathrm{by}133. \\ \mathrm{Step}1: \\ P\left(1\right)={11}^{1+2}+{12}^{2+1}=1331+1728=3059 \\ It\mathrm{is}\mathrm{divisible}\mathrm{by}133. \\ \mathrm{Step}2: \\ \mathrm{Let}P\left(m\right)\mathrm{be}\mathrm{divisible}\mathrm{by}133. \\ Now, \\ {11}^{m+2}+{12}^{2m+1}\mathrm{is}\mathrm{divisible}\mathrm{by}133. \\ \mathrm{Suppose}: \\ {11}^{m+2}+{12}^{2m+1}=133\lambda ...\left(1\right) \\ \mathrm{We}\mathrm{shall}\mathrm{show}\mathrm{that}P\left(m+1\right)\mathrm{is}\mathrm{true}\mathrm{whenever}P\left(m\right)\mathrm{is}\mathrm{true}. \\ \mathrm{Now}, \\ P\left(m+1\right)={11}^{m+3}+{12}^{2m+3} \\ ={11}^{m+2}.11+{12}^{2m+1}.{12}^2+11.{12}^{2m+1}-11.{12}^{2m+1} \\ =11\left({11}^{m+2}+{12}^{2m+1}\right)+{12}^{2m+1}\left(144-11\right) \\ =11.133\lambda +{12}^{2m+1}.133\left[\mathrm{From}\left(1\right)\right] \\ =133\left(11\lambda +{12}^{2m+1}\right) \\ It\mathrm{is}\mathrm{divisible}\mathrm{by}133. \\ \mathrm{Thus},P\left(m+1\right)\mathrm{is}\mathrm{true}. \\ Bythep\mathrm{rinciple}\mathrm{of}m\mathrm{athematical}i\mathrm{nduction},P\left(n\right)\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{all}\mathrm{n}\in N. \end{gathered}$$