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Mathematical Applications for the Management, Life, and Social Sciences11th Edition

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Chapter 13.1, Problem 25E
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Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Solutions

Chapter
Section
Problem 1CP
Problem 2CP
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
BuyFindarrow_forward

Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

In Problems 20-25, use the sum formulas I-V to express each of the following without the summation symbol. In Problems 20-23, find the numerical value.

25.   ∑ i = 1 n ( 1 − 2 i n + i 2 n 2 ) ( 3 n )

To determine

To calculate: The expression of ∑i=1n(1−2in+i2n2)(3n) without the summation symbol.

Explanation

Given Information:

The provided sum is ∑i=1n(1−2in+i2n2)(3n).

Formula used:

The value of the sum ∑k=1n(xk±yk)=∑k=1nxk±∑k=1nyk.

The value of the sum ∑k=1nk=n(n+1)2.

The value of the sum ∑k=1n1=n.

The value of the sum ∑k=1ncxk=c∑k=1nxk.

The value of the sum ∑k=1nk2=n(n+1)(2n+1)6.

Calculation:

Consider the sum ∑i=1n(1−2in+i2n2)(3n).

This can be written as,

∑i=1n(1−2in+i2n2)(3n)=∑i=1n(3n−3n⋅2in+3n⋅i2n2)=∑i=1n(3n−6in2+3i2n3)

Recall that the value of the sum ∑k=1n(xk±yk)=∑k=1nxk±∑k=1nyk.

Thus,

∑i=1n(1−2in+i2n2)(3n)=∑i=1n(3n−6in2+3i2n3)=∑i=1n3n−∑i=1n(6in2)+∑i=1n(3i2n3)

Recall that the value of the sum ∑k=1ncxk=c∑k=1nxk.

Thus,

∑i=1n(1−2in+i2n2)(3n)=∑i=1n3n−∑i=1n(6in2)+∑i=1n(3i2n3)=3n∑i=1n1−6n2∑i=1ni+3n3∑i=1ni2

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