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Q15E

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Discrete Mathematics and its Applications
Found in: Page 421
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

Show that (nk)2n for all positive integers n and all integers k with 0kn .

For all positive integers n and all integers k it's proven that nk2n .

See the step by step solution

Step by Step Solution

Step 1: Use Binomial theorem

Binomial theorem: binomial theorem, statement that for any positive integer n , the nth power of the sum of two numbers x and y may be expressed as the sum of n + 1 terms of the form.

(x+y)n=j=0n(nj)xnjyj

To proof:

nk2n

Step 2: For all positive integers  and all integers  with 0≤k≤n

Let n be a positive integer and an k integer with 0kn

2n=(1+1)n =j=0nnj(1)nj(1)j=j=0nnj(1)(1) =j=0nnj nk

Thus, .nk2n

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