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Compute the sum and the limit of the sum as $n \rightarrow \infty.$$$\sum_{i=1}^{n} \frac{1}{n}\left[4\left(\frac{2 i}{n}\right)^{2}-\left(\frac{2 i}{n}\right)\right]$$

$$\frac{13}{3}$$

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Integration

Section 2

Sums and Sigma Notation

Integrals

Integration Techniques

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

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But this it was like by multiplying in this one over end, on multiplying out beat this power to So we have the summation of Isaac to one and of we can get that as for and squared. And then, actually, that's cute because it was squaring this. And then we have two scored, which is four. So we have four times four. So it's not 16 and a nice word and then minus two over ence word I So I said this time here in this term here is a constant so we can put out on Let's start by breaking this off into two summations. So we have 16 over ANC, you of I squared I Z one to Andi have minus two over and swear information from one on of I. So looking at, there are 2.1. This is Equation three, and this is equation too. Okay, so we're rewriting them with their respective equation. This becomes 16 over an que challenge. I lost him and and plus one to end, plus one over six. And then we have minus two over end squared times and times on plus one. Okay, so let's simplify of it. So we see that we have 16/6. Um, that's equal to a two times three six of 8 8/3 two of eight. And then we cancel out one of these terms, and then we have to and squared. Plus bien, what's one over three times? All right. Okay. And then we have minus. We can cancel one of the students, and then we have minus two and plus two over, and okay, now let's combine this into a single fractions. Almost four, this time by three. Okay, So its walls, what out? Eight year. So eight times two. That's equal to 16 and squared. Plus three times. It's if you go to 24 and eight over three and squared. And then, well, two of minus three terms to which is six and squared minus six. And okay, so let's simplify this. So this 16 this is 16 minus six. So we get 10 and sway and in 24 minus six, that's plus 18 and plus eight over to be and squared. OK, so this is our fall information on. Now let's find the limited. This so knows that we have our hash power is and squared, so we'll divide by and squared in our numerator and denominator. We have the limit as un approaches. Infinity of 10 plus 18 over an was a spillover and squared over three to these council, and we're left with a 10 open three for our limits.

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