155 This is a very useful method to fnd a general formula for a sequence defned recursively as above. For example, consider the sequence a1 = 1; a2 = 1; an+2 = an+1 + 2an for n ∈ N. Solving the equation x2 = x+ 2 yields two solutions x 1 = 2 and x2 =(−1). Thus, + c2(−1) n x n = c12 n , where c1 and c2 are constants such as c1(2)+ c2(−1)= 1; c1(2) 2 + c 2(−1) 2 = 1. It is not hard to see that c1 = 1/3 and c2 = −1/3. Therefore, 1 1 2n − an = (−1) n for all n ∈ N. 3 3 Exercise 1.3.7. Hint: Prove frst that, for k = 1,2,...,n, we have n n n + 1 + = . k k − 1 k SECTION 1.4 Exercise 1.4.7. 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|.
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