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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 < 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.
 Fall '11
 math

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