Prove the following by the principle of mathematical induction:`\ 11^(n+2)+12^(2n+1)` is divisible 133 for all `n in Ndot`

Explanation

Let `P(n)=11^(n+2)+12^(2n+1)`
Step I For `n=1`,
`P(1) =11^(1+2)+12^(2xx1+1)=11^(3)+12^3`
`=(11+12)(11^2-11xx12+12^2)`
`=23 xx13` , which is divisible by 133.
Therefore , the result is true for `n=1`.
Step II Assume that the result is true for `n=k`, then
`P(k)=11^(k+2)+12^(2k+1)` is divisible by 133.
`rArr P(k)=133r`, where r is an integer .
Step III For `n=k+1`.
`therefore P(k+1)=11^((k+1)+2)+12^(2(k+1)+1)=11^(k+3)+12^(2k+3)`
`=11^((k+1)+1).11+12^(2k+1).12^2`
`=11.11^(k+2)+144.12^(2k+1)`

`therefore 11.11^((k+2))+144.12^(2k+1)=11(11^(k+2)+12^(2k+1))+133.12^(2k+1)`
i.e., `P(k+1)=11P(k)+133.12^(2k+1)`
But we know that , P(k) is divsible by 133. Also , `133.12^(2k+1)` is divisible by 133.
Hence , `P(k+1)` is divisible by 133. This shows that , the result is true for `n=k+1`.
Hence , by the principle of mathematical induction , the result is true for all . `n in N`.