Pearson eText for Probability & Statistics for Engineers and Scientists with R -- Instant Access (Pearson+)
Pearson eText for Probability & Statistics for Engineers and Scientists with R -- Instant Access (Pearson+)
1st Edition
ISBN: 9780137548552
Author: Michael Akritas
Publisher: PEARSON+
Question
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Chapter 2.3, Problem 19E
To determine

Expand (a12+2a2+a3)3 using multinomial theorem.

Expert Solution & Answer
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Explanation of Solution

Calculation:

Multinomial theorem:

The multinomial theorem for all the non-negative integers n1,n2,...nk is,

(a1+...+ar)n=n1+...+nr=n(nn1,n2,...,nr)a1n1a2n2...arnr

In the formula, (nn1,n2,...,nr)=n!n1!n2!...nr!

Consider,

(a12+2a2+a3)3=[(30,0,3)(a12)0(2a2)0(a33)+(30,1,2)(a12)0(2a2)1(a32)+(30,2,1)(a12)0(2a2)2(a31)+(30,3,0)(a12)0(2a2)3(a30)+(31,0,2)(a12)1(2a2)0(a32)+(31,1,1)(a12)1(2a2)1(a31)+(31,2,0)(a12)1(2a2)2(a30)+(32,0,1)(a12)2(2a2)0(a31)+(32,1,0)(a12)2(2a2)1(a30)+(33,0,0)(a12)3(2a2)0(a30)]

=[3!0!0!3!(1)(1)(a33)+3!0!1!2!(1)(2a2)(a32)+3!0!2!1!(1)(4a22)(a31)+3!0!3!0!(1)(8a23)(1)+3!1!0!2!(a12)(1)(a32)+3!1!1!1!(a12)(2a2)(a31)+3!1!2!0!(a12)(4a22)(1)+3!2!0!1!(a14)(1)(a31)+3!2!1!0!(a14)(2a2)(1)+3!3!0!0!(a16)(1)(1)]

=[a33+3(2a2)(a32)+3(4a22)(a3)+(8a23)+3(a12)(a32)+6(a12)(2a2)(a3)+3(a12)(4a22)+3(a14)(a31)+3(a14)(2a2)+(a16)]=a33+6a2a32+12a22a3+8a23+3a12a32+12a12a2a3+12a12a22+3a14a31+6a14a2+a16

Hence, the term (a12+2a2+a3)3 is expanded as,

a33+6a2a32+12a22a3+8a23+3a12a32+12a12a2a3+12a12a22+3a14a31+6a14a2+a16

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