Chapter 5.2, Problem 15E

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Using Sigma Notation In Exercises 11-16, use sigma notation to write the sum. [ ( 2 n ) 3 − 2 n ] ( 2 n ) + ⋯ + [ ( 2 n n ) 3 − 2 n n ] ( 2 n )

To determine

To calculate: The sum [(2n)32n](2n)++[(2nn)32nn](2n) in the form of sigma notation.

Explanation

Given: The provided sum is: [(2n)3âˆ’2n](2n)+â€¦+[(2nn)3âˆ’2nn](2n).

Formula used: The sum of n terms a1,a2,a3,â€¦,an using sigma notation is written as:

âˆ‘i=knai=ak+a2+a3+â€¦+an

[(2n)3âˆ’2n](2n)+â€¦+[(2nn)3âˆ’2nn](2n)

To find the sum in the form of sigma notation, apply the formula:

âˆ‘i=knai=ak+a2+a3+...+an

Where, i is the index of summation, n is the upper bound and k is the lower bound.

It can be noted that only index of summation is changing in any sigma notation. Therefore, in the

provided sigma notation changing term is index of summation. Upper bound is the highest value

of indexof summation in the provided sum and lower bound is the lowest value of index of

summation in the provided sum.

Therefore,

Upper bound, n=n.

Lower bound, k=1

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