n ) : + + + +…+ n = )( ) n n For n = , P( ): = ( )( ) × + = × × = which is true. Assume that P( k ) is true for some positive integer k, i.e., + + + +…+ k = )( ) k k k ... ( ) We shall now prove that P( k + ) is also true. Now, we have ( + + + +… +k ) + ( k + ) )( ) ) k k k k [Using ( )] )( ) ( ) k k k k )( ) k k k = ( )( ){ ( ) } k k k + + Thus P( k + ) is true, whenever P ( k ) is true.
Hence, from the principle of mathematical induction, the statement P( n ) is true for all natural numbers n . MATHEMATICS Example Prove that n > n for all positive integers n . Solution Let P( n ): n > n When n = , > . Hence P( ) is true.
Assume that P( k ) is true for any positive integer k , i.e., k > k ... ( ) We shall now prove that P( k + ) is true whenever P( k ) is true. Multiplying both sides of ( ) by , we get . k > k i.e., k + > k = k + k > k + Therefore, P( k + ) is true when P( k ) is true.
Hence, by principle of mathematical induction, P( n ) is true for every positive integer n . Example For all n ≥ , prove that ... . .
. ) n n Solution We can write P( n ): ... . .
. ) n n We note that