Let P(n)=2⋅42n+1+33n+1 Then P(1)=2⋅43+34=209, which is divisible by 11 but not divisible by 2,7 or 27 . Further, let P(k)=2⋅42k+1+33k+1 is divisible by 11 , that is, 2⋅42k+1+33k+1=11q for some integer q. Now P(k+1)=2⋅42k+3+33k+4 =2⋅42k+1⋅42+33k+1⋅33 =16⋅2⋅42k+1+27⋅33k+1 =16⋅2⋅42k+1+(16+11)⋅33k+1 =16[2⋅42k+1+33k+1]+11⋅33k+1 =16⋅1lq+11⋅33k+1 =11(16q+33k+1)=11m where m=16q+33k+1 is another integer. ∴P(k+1) is divisible by 11 .