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

Question

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

Graph showing points for sequence a_n in college algebra, chapter on sequences.

Graph showing points for sequence b_n in college algebra, chapter on sequences.

5Σi=4 (a_i/b_i)^3

Accepted Answer

Hello, today we're going to be using the two given graphs to evaluate the given summation. Now, before we take a look at the summation, let's go ahead and take a look at the two graphs. The graph on the lift lifts all the values in the sequence A sub N and the graph on the right lists all the values of the sequence B sub N. So what we want to go ahead and do is label every point that's given to us on the two graphs. And the way that we do that is we take a look at every end value and see what the corresponding a of end value is going to be. So for example, the first point of the sequence A sub N occurs when N is equal to one and that's going to give us a return value of positive four. So when N is equal to one, A sub N is equal to four, therefore, a sub one is going to equal to positive four. And we're going to repeat this process for the rest of the points. So when N is equal to two, we get the return value of positive one. So a sub two is equal to one. When N is equal to three, we get the return value of negative two. So a sub three is equal to negative two. And finally, when N is equal to four, we get the return value of negative five. So a sub four is equal to negative five. Now we're going to repeat this process for the B sequence. When N is equal to one, we get the return value of negative five. So B sub one is equal to negative five when N is equal to two, B is equal to negative two. So B sub two is equal to negative two. When N is equal to three, we get one. So B sub three is equal to positive one. And finally, when N is equal to four, we get positive four. So B sub four is equal to positive four. Now that we have all the points labeled, let's go ahead and take a look at the given summation. So what this is stating is that we want to add the first four terms of the B terms divided by the eight terms, those quantities cubed. So the way that we're gonna write this out is we're going to write out the first term when we have B sub one and a sub one. So we're going to be taking B sub one divided by a sub one which is going to be negative five divided by four. And that quantity is going to be cubed. Then we're going to repeat this process for B sub two and A sub two. So what we have is B sub two divided by a sub two which is going to be negative two divided by one that quantity cubed. Next we're going to have B sub three divided by a sub three which is going to be one divided by negative two that quantity cubed. Then finally, we're going to have B sub four divided by a sub four, which is going to be four divided by negative five, that quantity cubed. Okay. Now we have a bit of simplification to do Using the rule for exponential distribution, we can distribute these exponents to every numerator and denominator. So every numerator and denominator in the given sequence is going to be cubed and cubing, each value is going to give us negative over 64 -8/ -1/8 - for over 125. Now, if we want to add these values together, we're going to need to have the same denominator in each of the fractions. So we're going to need to look for the lowest common denominator. So in order to get this, we need to think what is going to be the least common multiple between 18 and 125. While in this case here, the least common multiple is going to be 8000. So we're going to need to multiply each fraction by a value that can give the denominator the value of 8000. And what we're going to do is multiply 100 negative 125 over 64 by 1 25/1 25. Then we're gonna multiply negative 8/1 by 8000 over 8000. Next, we're gonna multiply negative 1/8 by 1000 over 1000. And finally, we're multiplying negative 64/1 25 by 64/64. So if we multiply the first fraction by 1 25/1 we're going to get negative 15, over 8000. If we multiply the second fraction by 8000 over 8000, we get negative 64,000 over 8000, Multiplying the third fraction by 1000 over 1000 is going to give us negative 1000 over 8000. And finally multiplying the last fraction by 64/ is going to give us negative 4,096 over 8000. Now that we have the same denominator in each of the fractions, we can go ahead and combine this into a single fraction by combining all the enumerators together. And so the way that's gonna look Is we're going to have the numerator which is negative 15, -64,000 -1000 -4,096, which is going to give us a final total of negative 84,721 over 8000. And this is going to be the some nation simplified. And with that being said, the answer to this problem is going to be a 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.

5Σi=4 (a_i/b_i)^3

Key Concepts

Sequences

Summation Notation

Rational Functions

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