Evaluate the following summation: ∑ i = 0 4 2 i + 1 \sum_{i=0}^42i+1 ∑ i = 0 4 2 i + 1

give this problem a try. Here we're asked to evaluate the following summation as gonna be the sum from I equals zero to four of this function right here. Now right off the bat for this sum, I notice that we have two pieces being added together in this function. And recall that we can take this and write this as two separate sums. So we're going to have the sum from I equals zero to two, or excuse me, from I equals zero to four of this first piece to I, and it's gonna be plus, and then we're gonna have the sum from I equals zero to four of this next piece, which is one. So all we did was take this and split it into two separate sums. Now next what I'm going to use is the constant multiple rule. It says I can take any constants multiplied by our I variable here, and I can move them out front. So doing that, we're gonna take this two and move it to the front of the sum, and the rest of everything for the sum stays the same, so we'll still have I inside and then we won't have plus, and that's gonna be the sum from I equals zero to four of one. Now what I'm first going to do is evaluate this sum right here. So I'm gonna do that over to the side. So we're gonna have the sum from I equals zero to four of I. Now doing this, all I need to do is increment up by one each time starting from zero, getting all the way to four. And then we need to add all of our results. So since we just have I, we're gonna start with zero, that's gonna be plus one, plus two, plus three, plus four, and we have expanded out the sum here. So let's go ahead and simplify. Now one plus two is gonna give us three, then we have plus three, which gives us six, and then six plus four is 10, plus zero, that's all gonna come out to 10. So this is the result that we get for our sum. So we're gonna end up with two multiplied by 10 as this result. Now we'll deal with this next portion of the summation right here, which is the summation from I equals zero to four of the constant one. Now you may recall a rule that says that if we have the summation where we have some constant, we can just take that constant and multiply it by the high bound of four. But you might notice something here. This sum is a little bit different. It's starting at zero rather than starting at one. If it was started at one, that would be true. We could just multiply this by the front. Since it's starting at zero, well, we need to expand this. So what we would have is we'd start from zero, we go one, two, three, four. So then we'd have zero, and then that'd be one, and that'd be two, and that would be three, and then that would be four. So we have one plus one, plus one, plus one, plus one, which is five. Because notice that if I'm incrementing and starting from zero, I would in essence get five of this function being added together. It's just the function is constant. It would just be one. So it'd be one plus one plus one plus one plus one, which comes up to five, meaning this whole thing actually comes out to five. That's something you wanna watch out for. Right? You wanna watch out for whatever this bottom bound is, and you wanna make sure that that is something that you can actually use the rule for. So if it starts at one, you could multiply it. But since it starts at zero, we had to add this all out. And you can use this strategy every single time to make sure that you get the right answer, and you will get to whatever the correct solution is. So in this case, we're gonna have two times 10 plus five, which all comes out to 25, which is the solution to our problem. So that is how you can solve this sum. Let's move on to the next one.

8. Definite Integrals

Riemann Sums

Business Calculus

8. Definite Integrals

Riemann Sums

Struggling with Business Calculus?

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Multiple Choice

Evaluate the following summation:
$\sum_{i=0}^42i+1$

Verified step by step guidance

Understand the problem: The summation notation ∑_{i=0}^4 (2i + 1) means we need to calculate the sum of the expression (2i + 1) for each integer value of i from 0 to 4, inclusive.

Substitute each value of i (from 0 to 4) into the expression (2i + 1). For example, when i = 0, the term becomes 2(0) + 1 = 1. Repeat this for i = 1, 2, 3, and 4.

Write out the individual terms of the summation: (2(0) + 1), (2(1) + 1), (2(2) + 1), (2(3) + 1), and (2(4) + 1).

Add the results of these terms together to compute the total sum. This involves basic arithmetic addition of the values obtained in the previous step.

Verify your work by double-checking each substitution and addition to ensure accuracy in the final summation.