Explanation Given: ( x 1 + x 2 + x 3 + ............. + x m ) n = ∑ n 1 + n 2 + .... + n m = n c ( n : n 1 , n 2 , ...... , n m ) x 1 n 1 , x 2 n 2 ....... x m n m Where c ( n : n 1 , n 2 , ...... , n m ) = n ! n 1 ! n 2 ! .... m ! . Concept used: Principle of mathematical induction is a powerful mathematical principle to prove mathematical expression and formula. Here, first we assume, result is true for k and if the result also satisfy for k + 1 then it holds true for all. ( x + y ) n = ∑ r = 0 n n c r x n − r y r ( x 1 + x 2 + x 3 + ............. + x m ) n = ∑ r 1 + r 2 + .... + r m = 0 { c ( n 1 r 1 ) . c ( n − r 1 − r 2 , r 3 ) .......... c ( n − r 1 − r 2 ........ r m − 1 , r m ) } x 1 n − r 1 − r 2 ....... r m − 1 . x 2 r 2 . x 3 r 3 x 4 r 4 .......... x m r m Proof: Suppose the result is true for k = m then ( x 1 + x 2 + x 3 + ............. + x m ) n = ∑ r 1 + r 2 + .... + r m = 0 { c ( n 1 r 1 ) . c ( n − r 1 − r 2 , r 3 ) .......... c ( n − r 1 − r 2 ........ r m − 1 , r m ) } x 1 n − r 1 − r 2 ....... r m − 1 . x 2 r 2 . x 3 r 3 x 4 r 4

8th Edition

ISBN: 9781265370749

Author: ROSEN

Publisher: MCG

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Question

Chapter 6.5, Problem 65E

To determine

To prove:

If n is a positive integer, then (x1+x2+x3+.............+xm)n=∑n1+n2+....+nm=nC(n:n1,n2,......,nm)x1n1,x2n2.......xmnm.

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Chapter 6.5, Problem 64E

Chapter 6.5, Problem 66E

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The height of the graph of the probability density function f(x) varies with X as follows (round to four decimal places): X 16 Height of the Graph of the Probability Density Function You are flying out of Terminal 3 at JFK on a Wednesday afternoon between 3:00 and 4:00 PM. You get stuck in a traffic jam on the way to the airport, and if it takes you longer than 12 minutes to clear security, you'll miss your flight. The probability that you'll miss your flight is You have arrived at the airport and have been waiting 10 minutes at the security checkpoint. Recall that if you spend more than 12 minutes clearing security, you will miss your flight. Now what is the probability that you'll miss your flight? ○ 0.5 O 0.25 ○ 0.8333 ○ 0.6667

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Chapter 6 Solutions

DISCRETE MATH CONNECT ACCESS

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Knowledge Booster

To prove: If n is a positive integer, then ( x 1 + x 2 + x 3 + ............. + x m ) n = ∑ n 1 + n 2 + .... + n m = n C ( n : n 1 , n 2 , ...... , n m ) x 1 n 1 , x 2 n 2 ....... x m n m .

Solution Summary: The author explains that mathematical induction is a powerful mathematical principle to prove mathematical expression and formula.

To prove:

If n is a positive integer, then (x1+x2+x3+.............+xm)n=∑n1+n2+....+nm=nC(n:n1,n2,......,nm)x1n1,x2n2.......xmnm.

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Chapter 6 Solutions

To prove: If n is a positive integer, then ( x 1 + x 2 + x 3 + ............. + x m ) n = ∑ n 1 + n 2 + .... + n m = n C ( n : n 1 , n 2 , ...... , n m ) x 1 n 1 , x 2 n 2 ....... x m n m .

Solution Summary: The author explains that mathematical induction is a powerful mathematical principle to prove mathematical expression and formula.

To prove: If n is a positive integer, then (x1+x2+x3+.............+xm)n=∑n1+n2+....+nm=nC(n:n1,n2,......,nm)x1n1,x2n2.......xmnm.

Want to see the full answer?

Chapter 6 Solutions

To prove:

If n is a positive integer, then (x1+x2+x3+.............+xm)n=∑n1+n2+....+nm=nC(n:n1,n2,......,nm)x1n1,x2n2.......xmnm.