Introduction to Mathematical Analysis I - Second Edition

145 This is a very useful method to find a general formula for a sequence defined recursively as above. For example, consider the sequence a 1 = 1; a 2 = 1; a n + 2 = a n + 1 + 2 a n for n ∈ N . Solving the equation x 2 = x + 2 yields two solutions x 1 = 2 and x 2 = ( − 1 ) . Thus, x n = c 1 2 n + c 2 ( − 1 ) n , where c 1 and c 2 are constants such as c 1 ( 2 )+ c 2 ( − 1 ) = 1; c 1 ( 2 ) 2 + c 2 ( − 1 ) 2 = 1 . It is not hard to see that c 1 = 1 / 3 and c 2 = − 1 / 3. Therefore, a n = 1 3 2 n − 1 3 ( − 1 ) n for all n ∈ N . Exercise 1.3.8 . Hint: Prove first that, for k = 1 , 2 , . . . , n , we have n k + n k − 1 = n + 1 k . SECTION 1.4 Exercise 1.4.5 . In general, to prove that | a | ≤ m , where m ≥ 0, we only need to show that a ≤ m and − a ≤ m . For any x , y ∈ R , | x | = | x − y + y | ≤ | x − y | + | y | , This implies | x | − | y | ≤ | x − y | . Similarly, | y | = | y − x + x | ≤ | x − y | + | x | , This implies − ( | x | − | y | ) ≤ | x − y | . Therefore, || x | − | y || ≤ | x − y | .