= + + = = + + + , etc. It is worth to be noted that the sum of the first two odd natural numbers is the square of second natural number, sum of the first three odd natural numbers is the square of third natural number and so on.Thus, from this pattern it appears that + + + + ... + ( n – ) = n , i.e, the sum of the first n odd natural numbers is the square of n . Let us write P( n ): + + + + ...
+ ( n – ) = n . We wish to prove that P( n ) is true for all n . The first step in a proof that uses mathematical induction is to prove that P ( ) is true. This step is called the basic step.
Obviously = , i.e., P( ) is true. The next step is called the inductive step . Here, we suppose that P ( k ) is true for some PRINCIPLE OF MATHEMATICAL INDUCTION positive integer k and we need to prove that P ( k + ) is true. Since P ( k ) is true, we have + + + + ...
+ ( k – ) = k ... ( ) Consider + + + + ... + ( k – ) + { ( k + ) – } ... ( ) = k + ( k + ) = ( k + ) [Using ( )] Therefore, P ( k + ) is true and the inductive proof is now completed.
Hence P( n ) is true for all natural numbers n . Example For all n ≥ , prove that + + + +…+ n = ( )( ) n n Solution Let the given statement be P( n ), i.e., P(