155 This is a very useful method to find a general formula for a sequence defined 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, xn =c12 n +c 2(−1) n, where c1 andc2 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, an = 1 3 2n − 1 3 (−1)n for all n∈N. Exercise 1.3.7. Hint: Prove first that, for k =1,2, . . . ,n, we have n k + n k−1 = n+1 k . SECTION 1.4 Exercise 1.4.7. In general, to prove that |a| ≤m, where m≥0, we only need to show that a≤mand −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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