Give an example of a statement P ( n ) which is true for all n ≥ 4 but P ( 1 ) , P ( 2 ) and P ( 3 ) are not true. Justify your answer.

The required statement is P ( n ) = 2 n < n ! Justification P ( n ) : 2 n < n ! P ( 1 ) : 2 × 1 < 1 ! ⇒ 2 < 1 not true P ( 2 ) : 2 × 2 < 2 ! ⇒ 4 < 2 × 1 ⇒ 4 < 2 not true P ( 3 ) : 2 × 3 < 3 ! ⇒ 6 < 3 × 2 × 1 ⇒ 6 < 6 not true P ( 4 ) : 2 × 4 < 4 ! ⇒ 8 < 4 × 3 × 2 × 1 ⇒ 8 < 24 is true P ( 5 ) : 2 × 5 < 5 ! ⇒ 10 < 5 × 4 × 3 × 2 × 1 ⇒ 10 < 120 is true Hence, P ( n ) = 2 n < n ! is not true for P ( 1 ) , P ( 2 ) and P ( 3 ) but it is true for all values of n ≥ 4

Prove the following by the principle of mathematical induction: 1 + 2 + 3 + . . . + n = n n + 1 2 i.e, the sum of the first n natural numbers is n n + 1 2 .

Let us consider P ( n ) = 1 + 2 + 3 + - - - - - + n = n ( n + 1 ) 2 for n = 1 , 1 = 1 ( 1 + 1 ) 2 = 1 So p ( 1 ) is true now let us assume p ( k ) is true 1 + 2 + 3 . . . + k = k k + 1 2 is true we have to prove p ( k + 1 ) is true 1 + 2 + 3 . . + k + k + 1 = k ( k + 1 ) 2 + k + 1 = k ( k + 1 ) + 2 ( k + 1 ) 2 = ( k + 1 ) k + 2 2 = k + 1 k + 1 + 1 2 So, p ( k + 1 ) is true P ( 1 ) is true and P ( k + 1 ) is true whenever P ( k ) is true. Therefore, by the principle of mathematical induction, P ( n ) is true for all n ∈ N .

Prove the following by the principle of mathematical induction: 1 2 · 5 + 1 5 · 8 + 1 8 · 11 + . … . + 1 3 n - 1 3 n + 2 = n 6 n + 1

1 2 · 5 + 1 5 · 8 + 1 8 · 11 + . … . + 1 3 n - 1 3 n + 2 = n 6 n + 1 Let P ( n ) be the statement given by P n : 1 2 · 5 + 1 5 · 8 + 1 8 · 11 + . … . + 1 3 n - 1 3 n + 2 = n 6 n + 1 For n = 1 , L . H . S = 1 2 · 5 = 1 10 R . H . S . = 1 6 + 4 = 1 10 Since L.H.S. = R.H.S So, P ( 1 ) is true let P ( m ) be true = 1 2.5 + 1 5.8 + 1 8.11 + … + 1 3 m - 1 3 m + 2 = m 6 m + 4 ……(1) We wish to show that P (m + 1 ) is true for this we have to show that = 1 2.5 + 1 5.8 + 1 8.11 + … + 1 3 m - 1 3 m + 2 + 1 3 m + 1 - 1 3 m + 1 + 2 = m + 1 6 ( m + 1 ) + 4 L . H . S . = 1 2.5 + 1 5.8 + 1 8.11 + … + 1 3 m - 1 3 m + 2 + 1 3 m + 3 - 1 3 m + 3 + 2 = m 6 m + 4 + 1 3 m + 2 3 m + 5 ……. from (1) = m 2 ( 3 m + 2 ) + 1 3 m + 2 3 m + 5 = 1 3 m + 2 m 2 + 1 3 m + 5 = 1 3 m + 2 m 3 m + 5 + 2 2 3 m + 5 = 1 3 m + 2 3 m 2 + 5 m + 2 2 3 m + 5 = ( m + 1 ) 6 m + 10 = m + 1 6 m + 1 + 4 ⇒ P ( m + 1 ) is true Thus P m + 1 is true whenever P ( m ) is true Hence, by the principle of mathematical induction P ( n ) is true for all n ∈ N

Prove the following by the principle of mathematical induction: 1 1.4 + 1 4.7 + 1 7.10 + . … . + 1 3 n - 2 3 n + 1 = n 3 n + 1

1 1.4 + 1 4.7 + 1 7.10 + … + 1 3 n - 2 3 n + 1 = n 3 n + 1 Let P ( n ) be the statement given by P n : 1 1.4 + 1 4.7 + 1 7.10 + … + 1 3 n - 2 3 n + 1 = n 3 n + 1 For n = 1 , L . H . S = 1 1.4 = 1 4 R . H . S . = 1 3 × 1 + 1 = 1 3 + 1 = 1 4 Since, L . H . S = R . H . S So, P ( 1 ) is true let P ( m ) be true P m : 1 1.4 + 1 4.7 + 1 7.10 + … + 1 3 m - 2 3 m + 1 = m 3 m + 1 ………..(1) We wish to show that P ( m + 1 ) is true for this we have to show that 1 1.4 + 1 4.7 + 1 7.10 + … + 1 3 m - 2 3 m + 1 + 1 3 ( m + 1 ) - 2 3 ( m + 1 ) + 1 = m + 1 3 ( m + 1 ) + 1 L . H . S = 1 1.4 + 1 4.7 + 1 7.10 + … + 1 3 m - 2 3 m + 1 + 1 3 m + 1 3 m + 4 = 1 1.4 + 1 4.7 + 1 7.10 + … + 1 3 m - 2 3 m + 1 + 1 3 m + 1 3 m + 4 = m 3 m + 1 + 1 3 m + 1 3 m + 4 ….from (1) = 1 3 m + 1 m + 1 3 m + 4 = 1 3 m + 1 3 m 2 + 4 m + 1 3 m + 4 = 1 3 m + 1 m + 1 3 m + 1 3 m + 4 = m + 1 3 m + 4 = m + 1 3 m + 1 + 1 Thus P ( m + 1 ) is true for whenever P ( m ) is true Hence, by the principle of mathematical induction P ( n ) is true for all n ∈ N