   Chapter 13, Problem 1RE ### Mathematical Applications for the ...

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

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

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

# Calculate ∑ k = 1 8 ( k 2 + 1 ) .

To determine

To calculate: The value of the expression is k=18(k2+1).

Solution:

The value of the expression k=18(k2+1) is 212.

Explanation

Given Information:

The provided expression is k=18(k2+1).

Formula Used:

The sum of n numbers x1, x2, x3,,xn is given by

k=1nxk=x1+x2++xn

The value of the sum k=1n1 is n.

k=1n1=n

The sum of the square of first n natural numbers,

k=1nk2=n(n+1)(2n+1)6

Calculation:

Consider the sum k=18(k2+1).

Separate the components of the expression as follows:

k=18(k2+1)=k=18(k2)+k=18(1)

Apply the formula for the summation of the expressions,

Firstly, the sum of the first expression is:

k=1n(k2)=k(k+1)(2k+1)6k=18(k2)=8(9)(17)6=204

Secondly, the sum of the second expression is:

k=1n(1)=n.1k=18(1)=8.1=8

Add both the values for the expressions:

k=18(k2+1)=k=18(k2)+k=18(1)=204+8=212

Hence, the value of the expression k=18(k2+1) is 212.

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Evaluate the integral. 32. 13e3/xx2dx

Single Variable Calculus: Early Transcendentals

#### True or False: converges.

Study Guide for Stewart's Multivariable Calculus, 8th 