/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Will help you prepare for the ma... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Will help you prepare for the material covered in the first section of the next chapter. Evaluate \(j^{2}+1\) for all consecutive integers from 1 to 6 inclusive. Then find the sum of the six evaluations.

Short Answer

Expert verified
The sum of the evaluations of the expression \(j^{2}+1\) for the integers from 1 to 6 is 97.

Step by step solution

01

Substitution for \(j\)

We substitute the integers 1, 2, 3, 4, 5, and 6 for \(j\) in the expression \(j^{2}+1\). This results in the following six expressions: \(1^{2}+1\), \(2^{2}+1\), \(3^{2}+1\), \(4^{2}+1\), \(5^{2}+1\), and \(6^{2}+1\).
02

Evaluation of each expression

We evaluate each of the six expressions: \(1^{2}+1 = 2\), \(2^{2}+1 = 5\), \(3^{2}+1 = 10\), \(4^{2}+1 = 17\), \(5^{2}+1 = 26\), \(6^{2}+1 = 37\).
03

Summation of the results

We add up the results from step 2: \(2 + 5 + 10 + 17 + 26 + 37 = 97\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.