Evaluate the following summation: ∑ i = 1 2 4 i 3 − 3 i 2 + 2 i − 1 \sum_{i=1}^24i^3-3i^2+2i-1 ∑ i = 1 2 4 i 3 − 3 i 2 + 2 i − 1

This problem, we are given this polynomial here where we're asked to evaluate the following summation. Now right off the bat, I notice that we have a bunch of terms being added or subtracted, a bunch of pieces to this function. So what I'm going to do is break this up into multiple sums, where we find the sum of each individual piece being added or subtracted. So we'll have the sum from I equals one to two, that's gonna be a four I cubed. Then we'll have minus, then we'll have the sum from I equals one to two of three I squared, and then we're just gonna keep going. Then we'll have plus and then we'll have the sum from I equals one to two, and that's gonna be of two I. So it's gonna be this next portion. And then we'll have this final sum right here. So we'll have minus and then this sum, and it's gonna go from I equals one to two of just one. So notice how I kept these minus signs and these plus signs consistent, and I also kept each of these pieces consistent, I just split this up into multiple sums. Now from here, I'm going to use the constant multiple rule, which is I can take these constants and bring them out to the front of the sums. So doing this will have four, that's going to be multiplied by the sum from I equals one to two, and that's gonna be of I cubed. And we're just gonna keep doing this. So we'll have minus, and then we'll pull that three out, then we'll keep this sum the same, and that's gonna be for I squared. Then we're gonna have plus, and then we're gonna pull that two out. We'll keep this sum the same, and it's gonna be for just I. And then we'll keep this sum completely the same since I don't really need to pull this one out by itself since it's just the constant already by itself in the sum. So now we've gone ahead and made things a little bit more straightforward when evaluating these sums. So what I'm gonna do is go piece by piece to figure out what each of these sums are. We're gonna go ahead and start with this one right here. So we have the sum from I equals one to two of I cubed. And to solve this well, we start with one, so we won't have one cubed plus, and then we'll move up to two, two cubed. Now one cubed is one, two cubed is eight, one plus eight is nine. So that's what this result comes out to be. So going here, we're going to have four multiplied by nine. So this is what we get for the first piece. Now let's move on to this next portion. So we're gonna have minus this next portion, and I'm gonna go ahead and solve this one. Now solving this one, we'll have from I equals one to two of I squared, and that's going to be equal to and then we'll have one squared plus two squared. One squared is one, two squared is four, one plus four is five. So what we end up with is three multiplied by five. And that's going to be this next piece right here. So we have three times five, and then that's gonna be the solution to that portion. We're we're not done with the problem yet, but this is gonna be this piece evaluated. So now we're going to have plus, and then we'll move on to this sum. Now focus on this part right here, so we won't have the sum. I'll scroll down here so we have some more space. So the sum from I equals one to two of I. Now doing this, we'll have one plus two, which all comes out to three. This is gonna become two times three. And then we're going to have this last sum that we deal with, which is the sum from one to two of one. But notice here that our sum starts at one. Since it starts at one, what we can do is just take this two right here and multiply it by this constant. Since it's the sum of a constant, we'll have minus two times one, which is two. So two would be the result of this last sum, and that's going to be all of our numbers expanded out here. So we have nine times four, which comes up to 36. Then we'll have minus three times five, which is 15, plus two times three, which is six, minus two. If you go ahead and work out the math here, everything should come out to 25, which is the solution and result for this sum. So So that's how you can solve problems where you're dealing with these larger polynomials, and you need to find the sum. Hope you found this video helpful, and let's move on.

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:
$\sum_{i=1}^24i^3-3i^2+2i-1$

Verified step by step guidance

Step 1: Understand the problem. The summation notation ∑ indicates that we need to evaluate the sum of the expression 4i^3 - 3i^2 + 2i - 1 for all integer values of i from 1 to 6 (inclusive).

Step 2: Write out the terms of the summation. Substitute i = 1, 2, 3, 4, 5, and 6 into the expression 4i^3 - 3i^2 + 2i - 1 to generate the individual terms.

Step 3: Calculate each term. For example, when i = 1, the term is 4(1)^3 - 3(1)^2 + 2(1) - 1. Similarly, calculate for i = 2, i = 3, and so on up to i = 6.

Step 4: Add all the terms together. Once you have calculated the individual terms for each value of i, sum them up to get the total value of the summation.

Step 5: Verify your work. Double-check each substitution and calculation to ensure accuracy before finalizing the result.