In Exercises 61–68, use the graphs of and to find each indicat...

Question

In Exercises 61–68, use the graphs of and to find each indicated sum.

$\sum_{i = 1}^{5} (2 a_{i} + b_{i})$

Accepted Answer

Hello, today we're going to be using the given graphs to evaluate the specified summation. So before we take a look at the specified summation, let's go ahead and label the values on both the graphs. So the graph on the left represents the values for a sub N and the graph on the right represents the values for B sub N. So if, if we want to label the values here, we're going to take a look at when N is equal to one and the corresponding point is going to be here which gives us the Y value of A sub and value of negative two. So when N is equal to one, that means a sub one is going to equal to negative two. Continuing this pattern. When N is equal to two, we can see that a sub two is going to equal to zero. And if we continue this pattern, when N is equal to three, we get the value of two. So a sub three is going to equal to two. When N is equal to four, we get the value of four. So a sub four is going to equal to four when N is equal to five, we get the value of six. So a sub five is equal to six. And finally, when N is equal to six, we get the value of eight. So a sub six is equal to eight. Now we're gonna go ahead and label the points on the B graph. When N is equal to one on the B graph, we get the value of 10. So B sub one is equal to 10. When N is equal to two, we get seven. So B sub two is equal to seven. When N is equal to three, we get four. So B sub three is equal to four. When N is equal to four, we get the value of one. So B sub four is equal to one. When N is equal to five, we get negative two. So be sub five is equal to negative two. And finally, when N is equal to six, we get the value of negative five. So B sub six is equal to negative five. So now that we have all the terms listed on the graph, let's go ahead and take a look at the summation. So what this summation is saying is that we want the sum of all six terms of three times every term in the A graph minus five times every term in the B graph. So the way that we're gonna write this out is we're gonna go ahead and first list the term when N is equal to one, when N is equal to one, we get three times a sub one which was negative two minus five times B sub one which was positive 10. And we're going to repeat this pattern going all the way to an end is equal to six. So for the next term, when N is equal to two, we get three times A sub two which is zero minus five times B sub two, which is seven. Next, when N is equal to three, we get three times A sub three which is two minus five times B sub three, which is four, then we're going to add when N is equal to four. So we get three times a sub four which is four - times a species up for which is one plus when N is equal to five, we get three times a sub five which is six minus five times B sub five which is negative two. And I apologize, I'm running out of room here. But then we're gonna add the final term when N is equal to six. So that's going to give us three times A sub six which is eight minus five times B sub six which is negative five. Okay. Now that we have all the terms for the summation listed out. Let's go ahead and do some simplification in the first term. Three times negative two is going to give us negative six and negative five times 10 is going to give us negative 50. In the second term three times zero is going to give us zero and negative five times seven is going to give us negative 35. In the third term three times two is going to give us six and negative five times four is going to give us negative 20. In the fourth term four times three is going to give us 12 negative five times one is going to give us negative five. In the fifth term, six times three is going to give us 18 and negative five times negative two is going to give us positive 10. And finally in the last term, eight times three is going to give us 24 and negative five times negative five is going to give us positive 25. So simplifying the terms that was given to us what we now have is negative six minus 50 plus zero minus 35 which is going to give us negative 35 plus six minus Plus 12 -5 Plus plus 10 Plus 24 plus 25. And simplifying these terms further negative six minus 50 is going to give us negative 56. We still have negative 35 here. So we don't worry about that term. Six minus 20 is going to give us negative 14, 12 -5 is going to give us seven 18 plus 10 is going to give us 28 plus 25 is going to give us 49. So what we are left with is negative 56 minus minus 14 plus seven plus 28 plus 49. And adding these terms together is going to give us a final total of negative 21 which is going to be the summation evaluated. And with that being said, the answer to this problem is going to be c so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 61–68, use the graphs of and to find each indicated sum.

$\sum_{i = 1}^{5} (2 a_{i} + b_{i})$

Key Concepts

Sequences and Terms

Summation Notation

Combining Sequences in Sums

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