**Mathematical induction is a unique and special way to prove the things, in only two steps.***Step 1.* Show that it is true for n = 1.*Step 2.* Show that if n = k is true then n = k+1 is also true.**For example:****Prob. **By principal of mathematical induction prove that11^{n+2} + 12^{2n+1} is divisible by 133, n ∈ N.**Solution.***Step 1 P(1)- *Show it is true for n = 111^{n+2} + 12^{2n+1} = 11^{1+2} + 12^{2(1)+1} = 1331 + 1728 = 3059

Yes 3059 is divisible by 133.11^{1+2} + 12^{2(1)+1} is true.*Step 2 P(k)- *Assume it is true for n = k

11^{k+2} + 12^{2(k)+1} is true.

(above line is an assumption only, which we will use as a fact in rest of the solution)

Now, prove that 11^{(k+1)+2} + 12^{2(k+1)+1 }is divisible by 133. (here n = k +1 now, P(k+1))**We have,**

P(k+1)

11^{(k+1)+2} + 12^{2(k+1)+1} = 11^{k+3} + 12^{2k+3}

11^{(k+1)+2} + 12^{2(k+1)+1} = 11^{k+2} x 11 + 12^{2k+1}x 12^{2}

11^{(k+1)+2} + 12^{2(k+1)+1} = (11^{k+2} x 11) + (12^{2k+1}x 144 )

11^{(k+1)+2} + 12^{2(k+1)+1} = (11^{k+2} x 11) + (12^{2k+1}x (11 + 133))

11^{(k+1)+2} + 12^{2(k+1)+1} = (11^{k+2}+ 12^{2k+1})x 11+(12^{2n+1} x 133)

11^{(k+1)+2} + 12^{2(k+1)+1} = ((11^{k+2}+12^{2k+1})x 11)+(12^{2n+1} x 133)

Here 11^{k+2}+12^{2k+1} is divisible by 133 as assumed in n = k, P(1),

And 12^{2n+1} x 133 is multiple of 133 so it is divisible by 133.

So,

11^{(k+1)+2} + 12^{2(k+1)+1} = ((divisible by 133)x 11)+(divisible by 133)

11^{(k+1)+2} + 12^{2(k+1)+1} = divisible by 133.

In this problem

If n = n, i.e, P(1) is true then n = n+1, i.e, P(n+1) is also true. **Hence proved.**