In Exercises 29–42, find each indicated sum. 4Σi=1 (−1/2)^i

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Hello today we're going to be evaluating the summation of K is equal to 1 to 5 of negative 2/3 to the power of K. Now, before we start this I'm just gonna go ahead and quickly rewrite the pattern given to us -2/3 Is the same thing as -1 times 2/3. So rewriting the argument is going to give us negative one Times Times 2/3, raised to the power of K. And using our exponential distribution, we're going to distribute this exponent to each term of the product so we can rewrite the summation as the summation from K is equal 1-5 of negative one to the power of K times to over three to the power of K. Now let's go ahead and start evaluating the summation. And in order to start evaluating this we need to figure out what is the lowest term and the highest term for the index to plug into the pattern. The lowest term could be found at the bottom of sigma and in this case K is going to equal to one is going to be the starting value. Now we want to plug in all the values that come starting with one and come after one going to the final value which in this case is going to be five at the top of sigma. So we're going to be plugging in case equal to 1234 and five into our given pattern and we're going to take all of those terms and added together to figure out what the sum is going to be. So let's go ahead and start that process When K is equal to one, We get negative one To the power of one, times 2/3 to the power of one, -1 to the power of one is just going to give us - And to over three to the power of one is going to give us 2/3. So we have negative one times 2/3 and that's going to give us negative 2/3 as our first term. And we're going to repeat this process for the remaining values of K. Next, when we plug in K is equal to two, we get negative one to the power of two times 2/3 to the power of two. Negative one. Race to the even power is going to give us a positive value one so we can replace this with just one and for 2/3 raise to the power of two. We want to go ahead and distribute the expo into both the numerator and denominator And this is going to give us one times two squared which is 4/3 squared which is nine and one times 4/9 is just going to give us 4/9 which is going to be our second term. Next we're going to plug in K is equal to three and that's going to give us negative one to the power of three times 2/ To the power of three. Now negative one race to any odd power is going to give us -1. So we can replace this with negative one and once again we want to distribute the exporting of three to both the numerator and denominator. And that's going to give us negative one times two to the power of three. Which is going to give us 8/3 to the power of three which is going to give us 27 negative one times 8/27 is going to give us negative 8/ that's going to be the next term of our some Next we're gonna plug in K is equal to four when K is equal to four. We get negative one to the power of four times 2/ To the power of four. Again negative one. Race to any even value is going to give us positive one and we're going to distribute the X one and four to both the numerator and denominator. That's going to give us one times two to the power of four which is 16/3 to the power of four which is 16/81 times one is going to give us 16/81. And finally we're going to plug in the value K is equal to five and that's going to give us negative one to the power of five times 2/3 to the power of five negative 12. Any odd exponent is going to give us negative one and we're going to distribute the X one and five to both the numerator and denominator. That's going to give us negative one times two to the fifth power which is 32/3 to the fifth power which is And 32 over 243 times negative one is going to give us negative 32 over 243 and that's going to be the final term of the sum. Now what we want to go ahead and do is take each term that we calculated for and put it all together into a single sum. And that's going to give us negative 2/ Plus 4/9 -8/ Plus 16/ -32/243. And adding these values together is going to give us negative over 243 and that's going to be the total sum for the given summation. 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 evaluate a summation and I'll go ahead and see you all in the next video

In Exercises 29–42, find each indicated sum. 4Σi=1 (−1/2)^i

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