2. (a) For n ≥ 0 and x E R, prove that (b) Prove n n Σ ^ (^.) 2^* (1 - 2)"-* k k=0 = = In. n x)n-k Σk(k − 1) (7) a* (1 − 2)¹-* = 2²n(n − 1). - k=0 (c) Finally prove Lemma 6 in Lecture 3: Σ(k − nx)² (^.) x^(1 − a)¹-* = x(1 − x)n. - k k=0

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.4: Values Of The Trigonometric Functions
Problem 22E
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real analysis
1111
2. (a) For n > 0 and x € R, prove that
(b) Prove
n
Σ(*) (
k=0
n
x*(1 − x)n-k
Σκ« - 1)(*)*(1 – g)*-* = r'n(n - 1).
Σκ(κ
k=0
(c) Finally prove Lemma 6 in Lecture 3:
k=0
= τη.
Σ
(k – n
na) ² (1) 2² (1-2)^
|x^(1 − x)n-k = x(1 − x)n.
του -
Transcribed Image Text:1111 2. (a) For n > 0 and x € R, prove that (b) Prove n Σ(*) ( k=0 n x*(1 − x)n-k Σκ« - 1)(*)*(1 – g)*-* = r'n(n - 1). Σκ(κ k=0 (c) Finally prove Lemma 6 in Lecture 3: k=0 = τη. Σ (k – n na) ² (1) 2² (1-2)^ |x^(1 − x)n-k = x(1 − x)n. του -
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