summation

# Then n k1 fk fn2 1 induction basis for n 1 we

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Unformatted text preview: mbers Let f0 = 0 and f1 = 1 and fn = fn−1 + fn−2 for n ≥ 2. Then n ￿ k=1 fk = fn+2 − 1. Induction basis: For n = 1, we have 1 ￿ k=1 fk = 1 = (1 + 1) − 1 = f1 + f2 − 1 = f3 − 1 22 Sum of Fibonacci Numbers Induction Hypothesis: Let us assume that the claim holds for n − 1. Induction Step: Assuming the Induction Hypothesis, we will show that the claim holds for n. n ￿ fk = k=1 fn + n− 1 ￿ fk k=1 = = fn + fn+1 − 1 by Ind. Hyp. fn+2 − 1 by deﬁnition Therefore, the claim follows by induction on n. 23 Other Useful Sums n ￿ k=1 n(n + 1)(2n + 1) k= 6 n ￿ k=1 2 n2 (n + 1)2 3 k= 4 24 Infinite Geometric Series Let x be a real number such that |x| &lt; 1. Then ∞ ￿ k=0 1 x= . 1−x k 25 Infinite Geometric Series Since the sum of a geometric series satisﬁes n ￿ k=0 xn+1 − 1 xk = , x−1 we have ∞ ￿ k=0 xk = lim n→∞ n ￿ k=0 xn+1 − 1 xk = lim n→∞ x−1 As limn→∞ xn+1 = 0, we get ∞ ￿ k=0 −1 1 x= = . x−1 1−x k 26 Another Useful Sum ∞ ￿ kx k −1 k=1 1 = (1 − x)2 Diﬀerentiating both sides of ∞ ￿ k=0 yields the claim. 27 1 x= 1−x k...
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## This note was uploaded on 03/24/2014 for the course CSCE 222 taught by Professor Math during the Fall '11 term at Texas A&M.

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