In Exercises 8–9, find each indicated sum. This is a summation,...

Question

In Exercises 8–9, find each indicated sum.  This  is a summation, refer to the textbook.

Accepted Answer

Hello today we're going to be finding the sum of a given summation. So the summation given to us is the summation from Isaac to one of four of three. I squared minus five. Now a trick that we can do is to go ahead and separate this into two separate summations. Doing so is going to give me the summation for when I is equal to 14 of three, I squared minus the summation from when I is equal to one of four of five. Now on the right hand side this is just a summation with the constant as its argument. So when you have a constant as the argument of a summation you just multiply that by the end point of the summation and this is going to simplify to just five times four. Now on the left hand side you need to expand this from the start point to the end point. So you're going to expand this from one is equal one all the way to is equal to four. And doing so is going to give us three times one squared plus three times two squared plus three times three squared plus three times four squared. And we're subtracting five times four from this summation. Okay let's go ahead and simplify now One squared is just gonna be one and one times three is going to be three two squared is going to be 44 times three is going to give us 12, three squared is nine and nine times three is gonna give us and four squared is 16 16 times three is gonna give me 48 5 times four is 20. So we're gonna subtract 20 from this. So three plus 12 plus 27 plus minus 20 Is going to equal to 70. And that is gonna be the sum for our summation, which means that the solution to this problem is going to be d. So I hope this video helps you in understanding how to find the sum of a summation and I'll go ahead and see you all in the next video.

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In Exercises 8–9, find each indicated sum. This is a summation, refer to the textbook.

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