Prove that, for all integers n ≥ 0 and 0 ≤ k ≤n, we have (x+¹)=()+(*+¹) ++ (1) n+ k+1 k Hint: the left hand side is the number of k + 1-element subsets of {1,..., n+1}. Show that the right hand side is also equal to this number by grouping these subsets according to their largest element.
Prove that, for all integers n ≥ 0 and 0 ≤ k ≤n, we have (x+¹)=()+(*+¹) ++ (1) n+ k+1 k Hint: the left hand side is the number of k + 1-element subsets of {1,..., n+1}. Show that the right hand side is also equal to this number by grouping these subsets according to their largest element.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Prove that, for all integers n ≥ 0 and 0 ≤ k ≤ n, we have
k+
(+) = (3) + (*+¹) ++ (1)
Hint: the left hand side is the number of k + 1-element subsets of {1,...,n+1}. Show that the
right hand side is also equal to this number by grouping these subsets according to their largest
element.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F96191c65-e923-461d-9658-eb756d234e0f%2Fbd675a67-2607-43e5-8b5e-7b424e076923%2Fisexf69_processed.png&w=3840&q=75)
Transcribed Image Text:Prove that, for all integers n ≥ 0 and 0 ≤ k ≤ n, we have
k+
(+) = (3) + (*+¹) ++ (1)
Hint: the left hand side is the number of k + 1-element subsets of {1,...,n+1}. Show that the
right hand side is also equal to this number by grouping these subsets according to their largest
element.
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