Stitz-Zeager_College_Algebra_e-book

Generalizing this a bit we see that if we have a

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Unformatted text preview: oceeding as before, replacing in this case the variable k with the appropriate whole number, beginning (−1)0 (−1)1 (−1)2 (−1)3 1 with 0, we get b0 = 2(0)+1 = 1, b1 = 2(1)+1 = − 3 , b2 = 2(2)+1 = 1 and b3 = 2(3)+1 = − 1 . 5 7 (This sequence is called an alternating sequence since the signs alternate between + and −. The reader is encouraged to think what component of the formula is producing this effect.) 9.1 Sequences 553 3. From {2n − 1}∞ , we have that an = 2n − 1, n ≥ 1. We get a1 = 1, a2 = 3, a3 = 5 and n=1 a4 = 7. (The first four terms are the first four odd natural numbers. The reader is encouraged to examine whether or not this pattern continues indefinitely.) 4. Proceeding as in the previous problem, we set aj = 1 a4 = 2 and a5 = 0. 1+(−1)j , j j ≥ 2. We find a2 = 1, a3 = 0, 5. To obtain the terms of this sequence, we start with a1 = 7 and use the equation an+1 = 2 − an for n ≥ 1 to generate successive terms. When n = 1, this equation becomes a1...
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