Prove the following by using the principle of mathematical induction for all
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),
       Â
∴      Â
  Â
∴      P (m + 1) is true.
∴      P(m) is true. is true.
        Hence, by the principal of mathematical induction, P(n) is true for all
Use principle of mathematical induction to prove that:
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) =
                                      Â
                       [From (i), 1 + 2 + 3 + ...... + m = ]
∴       P (m + 1):
 Â
   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
     Â
Prove by mathematical induction that sum of cubes of three consecutive natural numbers is divisible by 9.
Let n, n+1, n+2 be three consecutive natural numbers.
Let P(n): is divisible by 9.
I. Â Â Â 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,Â
  P(m) : is divisible by 9.
                            ...(i)
III.  For n = m + 1,                          Â
      is divisible by 9.
      Now, from (i),
    Â
           Â
   Â
       Â
        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
Prove the following by using the principle of mathematical induction for all
a + (a + d) + (a + 2d) + ...........+ [a + (n - 1)d] =
Let P(n) : a + (a + d) + (a + 2d) + .............+ [a + (n - 1)d] =
I.    For n = 1,
    Â
 Â
∴     P(1) is true
II.      Suppose the statement is true for n=m,
∴      .... (i)
III.  For n = m + 1,
     Â
or   Â
    From (i),
   Â
∴    Â
          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
Â
Â
Prove the following by using the principle of mathematical induction for all .
Let
I.      For n = 2(note this step, n>1)
      Â
       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,
        Â
or    Â
or     Â
        Adding on both sides of (i), we get
Â
       Â
   Â
    Â
    Â
        But,
∴    Â
∴        P (m + 1) is true
∴        P(m) is true P(m + 1) is true
Hence, P(n) is true for all
          Â