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enumeration of terms with ellipses becomes
tedious.
18 Geometric Series
If a and r = 0 are real numbers, then
n+1
n
ar
−a
if r = 1
j
r −1
ar =
(n + 1)a if r = 1
j =0 Proof:
The case r = 1 holds, since arj = a for each of the
n + 1 terms of the sum.
The case r = 1 holds, since
n
n
n
(r − 1) j =0 arj =
arj +1 −
arj
= j =0
n+1
j =0 arj − j =1
n+1 n
arj j =0 = ar
−a
and dividing by (r − 1) yields the claim.
19 Sum of First n Positive Integers
For all n ≥ 1, we have
n
k = n(n + 1)/2
k=1 We prove this by induction.
Basis step: For n = 1, we have
1
k = 1 = 1(1 + 1)/2. k=1 20 Sum of the First n Positive Integers
Induction Hypothesis: We assume that the claim
holds for n − 1.
Induction Step: Assuming the Induction Hypothesis, we will show that the claim holds for n.
n k=1 k =
=
=
= n− 1 n + k=1 k
2n/2 + (n − 1)n/2 by Induction Hypothesis
2n+n2 −n
2
n(n+1)
2 Therefore, the claim follows by induction on n.
21 Sum of Fibonacci Nu...
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 Fall '11
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