/*! 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 24 Evaluate the sums in Exercises \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate the sums in Exercises \(19-28\). $$ \sum_{k=1}^{6}\left(k^{2}-5\right) $$

Short Answer

Expert verified
The sum is 61.

Step by step solution

01

Understanding the Sum

The expression to be evaluated is \[ \sum_{k=1}^{6}(k^2 - 5) \]This means we need to find the sum of the expression \(k^2 - 5\) for each integer \(k\) from 1 to 6.
02

Expanding the Series

Calculate the value of \(k^2 - 5\) for each \(k\):- For \(k = 1\): \(1^2 - 5 = -4\)- For \(k = 2\): \(2^2 - 5 = -1\)- For \(k = 3\): \(3^2 - 5 = 4\)- For \(k = 4\): \(4^2 - 5 = 11\)- For \(k = 5\): \(5^2 - 5 = 20\)- For \(k = 6\): \(6^2 - 5 = 31\)
03

Calculating the Sum

Now, add all the calculated values together:\[(-4) + (-1) + 4 + 11 + 20 + 31\]Step by step:- \(-4 + (-1) = -5\)- \(-5 + 4 = -1\)- \(-1 + 11 = 10\)- \(10 + 20 = 30\)- \(30 + 31 = 61\)
04

Final Answer

The sum of the series \(\sum_{k=1}^{6}(k^2 - 5)\) is 61.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Expansion
Series expansion is the process of writing out each term of a summation to make it easier to work with. In this case, we are given a summation expression: \( \sum_{k=1}^{6}(k^2 - 5) \). This instructs us to evaluate the expression \( k^2 - 5 \) for every integer value \( k \) from 1 to 6.
The series begins with the smallest value of \( k \), which is 1, and increases step-by-step up to 6. At each step, the expression \( k^2 - 5 \) is evaluated. By writing out each term explicitly, one can see the individual contributions to the overall sum.
  • For \( k = 1 \), the expression becomes \( 1^2 - 5 = -4 \).
  • For \( k = 2 \), the calculation yields \( 2^2 - 5 = -1 \).
  • Continuing in this way, the results are \( 4, 11, 20, \) and \( 31 \) for \( k = 3, 4, 5, 6 \) respectively.
By laying out each step clearly, series expansion helps in simplifying calculations and allows you to organize complex expressions into manageable parts.
Arithmetic Calculation
Once you have expanded the series and worked out each term, the next step is arithmetic calculation. This means adding up all those values you calculated in the series expansion.
The process involves basic arithmetic operations (in this case, addition) that are performed sequentially to arrive at a final sum. Let's review this calculation:
We have the following terms from our expansion:
  • \(-4\)
  • \(-1\)
  • \(4\)
  • \(11\)
  • \(20\)
  • \(31\)
Add these step-by-step:
  • Start by adding \(-4 + (-1)\) to get \(-5\).
  • Then add \(-5 + 4\) to reach \(-1\).
  • Next, calculate \(-1 + 11\) resulting in \(10\).
  • Proceed with \(10 + 20\) which gives \(30\).
  • Finally, adding \(30 + 31\) you arrive at \(61\).
Each calculation involves simple addition, staying organized makes it easier to avoid mistakes, ensuring the correct final sum.
Evaluating Algebraic Expressions
Evaluating algebraic expressions like \( k^2 - 5 \) involves substituting values and performing mathematical operations. Algebraic expressions can include constants, variables, and operations like addition, subtraction, multiplication, or division.

In the given exercise, the expression \( k^2 - 5 \) means:
  • \( k^2 \) is the square of the variable \( k \).
  • Subtract \( 5 \) from \( k^2 \).
When you substitute values for \( k \) (specifically, 1 through 6 in this case), you are solving the expression for each of those values. Calculating \( k^2 \) means multiplying \( k \) by itself, and the subtraction then adjusts the outcome according to the expression format.
This demonstrates a common task in algebra: plug in different values into an expression to get specific results, which are then summed up if required by the context of the problem, like in our summation. Understanding how to work with algebraic expressions is fundamental for solving many mathematical problems.

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

Use the Substitution Formula in Theorem 6 to evaluate the integrals in Exercises \(1-24 .\) $$ \text { a. }\int_{0}^{\sqrt{7}} t\left(t^{2}+1\right)^{1 / 3} d t \quad \text { b. } \int_{-\sqrt{7}}^{0} t\left(t^{2}+1\right)^{1 / 3} d t $$

Suppose that $$\int_{0}^{1} f(x) d x=3$$ Find $$\int_{-1}^{0} f(x) d x$$ if a. \(f\) is odd, \(\quad \mathbf{b} . f\) is even.

Find the areas of the regions enclosed by the lines and curves in Exercises \(51-58 .\) $$ x-y^{2 / 3}=0 \quad \text { and } \quad x+y^{4}=2 $$

In Exercises \(91-94\) , you will find the area between curves in the plane when you cannot find their points of intersection using simple algebra. Use a CAS to perform the following steps: a. Plot the curves together to see what they look like and how many points of intersection they have. b. Use the numerical equation solver in your CAS to find all the points of intersection. c. Integrate \(|f(x)-g(x)|\) over consecutive pairs of intersection values. d. Sum together the integrals found in part (c). $$ f(x)=x^{2} \cos x, \quad g(x)=x^{3}-x $$

A basic property of definite integrals is their invariance under translation, as expressed by the equation. $$\int_{a}^{b} f(x) d x=\int_{a-c}^{b-c} f(x+c) d x$$ The equation holds whenever \(f\) is integrable and defined for the necessary values of \(x\) . For example in the accompanying figure, show that $$\int_{-2}^{-1}(x+2)^{3} d x=\int_{0}^{1} x^{3} d x$$ because the areas of the shaded regions are congruent. (GRAPH NOT COPY) For each of the following functions, graph \(f(x)\) over \([a, b]\) and \(f(x+c)\) over \([a-c, b-c]\) to convince yourself that Equation (1) is reasonable. a. \(f(x)=x^{2}, \quad a=0, \quad b=1, \quad c=1\) b. \(f(x)=\sin x, \quad a=0, \quad b=\pi, \quad c=\pi / 2\) c. \(f(x)=\sqrt{x-4}, \quad a=4, \quad b=8, \quad c=5\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.