Let P(n):

I.         For n = 1,

           P(1) : 1 =

∴         P(1) is true

II.        Let the statement be true for n = m,

     P(m) : 1 + 3 + 5 + .................... + (2m - 1) = m²                    ...(i)

III.      For n = m + 1,

         P(m + 1) : 1 + 3 + 5 + .......... + [2 (m+1) - 1] = (m + 1)²

or      1 + 3 + 5 + ........... + (2m - 1) + (2m + 1) = (m + 1)²

        From (i),

        

∴       

   

∴       P (m + 1) is true.

∴       P(m) is true. is true.

         Hence, by the principal of mathematical induction, P(n) is true for all

Let

I.       For n = 2(note this step, n>1)

       



        which is true

∴       P(n) is true for n = 2

II.      Suppose the statement is true for n = m,

                         .... (i)

III.      For n = m + 1,

         

or     

or      

         Adding on both sides of (i), we get
 
        

    

     

     

         But,

∴     

∴         P (m + 1) is true

∴         P(m) is true P(m + 1) is true

Hence, P(n) is true for all

Let P(n): 1 + 2 + 3 + ......... + n =

I. For n = 1,

    P(1) : 1 = is true.

II.  Suppose the statement is true for n = m,

      i.e. P(m):           ....(i)

III.    For n = m + 1,

        P(m + 1): 1 + 2 + 3 + ........ + (m + 1) =

or  [1 + 2 + 3 + ...... + m] + (m + 1) =
                                       
                                         [From (i), 1 + 2 + 3 + ...... + m = ]

∴        P (m + 1):



  



    which is true

∴    P(m + 1) is true

∴    P(m) is true P(m + 1) is true

Hence, by mathematical induction

P(n) is true for all

Let n, n+1, n+2 be three consecutive natural numbers.

Let P(n): is divisible by 9.

I.       For n = 1,

       is divisible by 9

  1 + 8 + 27 is divisible by 9 36 is divisible by 9

      which is true

∴     the statement is true for n = 1.

II.     Suppose the statement is true for n = m, 

   P(m) : is divisible by 9.

                              ...(i)

III.   For n = m + 1,                           

       is divisible by 9.

       Now, from (i),
     
             

    

        

         where

     is divisible by 9

       P (m + 1) is true

∴         P (m) is true P (m + 1) is true.

           Hence, by the principal of mathematical induction, P (n) is true for all

Let P(n) : a + (a + d) + (a + 2d) + .............+ [a + (n - 1)d] =

I.     For n = 1,

     

  

∴      P(1) is true

II.       Suppose the statement is true for n=m,

∴       .... (i)

III.   For n = m + 1,

      

or      

       From (i),

       

∴     









           which is true

∴          P (m + 1) is true

∴          P (m) is true P(m + 1) is true

Hence by the principle of mathematical induction, P(n) is true for all