In Exercises 31–36, find the indicated sum. Use the formula for t...

Question

In Exercises 31–36, find the indicated sum. Use the formula for the sum of the first n terms of a geometric sequence. 10 Σ (i = 1) 5 · 2^i

Mathematical expression for the sum of a geometric sequence in college algebra.

Accepted Answer

everyone in this problem? We are asked to evaluate the specified sum for the given geometric sequence. And the some were given is the sum from of k equals 1 to 12. 4 times five to the k. Okay. All right, so let's think about this. It's the sum of a geometric sequence. We want to evaluate the sum Now are some is related to the general form of the sum of a geometric sequence which is given by the following the sum K. Well zero to n minus one of a. R to the exponent K. Okay. And what you'll notice Is that this general form starts at K equals zero But the somewhere given starts at Cake was one. Okay, so we can't use that general some formula if this is not in the same form. So we need to take our original some that were given some from K equals 1-12 of four times five to the K. And we need to rewrite this so that we have K starting at zero. How do we do this? Okay, well let's think about this. We know that we can factor or pull out a factor of five from our five to the exponent. Okay, okay, so we can write this as the sump from k equals 1 to 12. Of four times five Times five for the exponent K -1. Okay, these are equivalent. Okay, on the left hand side when k is one we get four times five. On the right hand side we get when k is one we get four times five times five to the exponent zero which is just one. So we get just four times five. Okay so these are equivalent now that we've shifted our exponents by one. This allows us to change that indexing and we can write this instead as the sum From let's use J. So that we don't confuse ourselves. Okay from zero We're shifting the bottom by one, we have to shift the top by one. So from 0 to 11 of four times 5 Times 5 to the exponent J. Because we've shifted it down. Alright and we can check this again on the left hand side when K is one, we get four times five times five to the exponent zero. So we get four times five on the right hand side we have J Zero. First we get four times five times five to the zero, Same thing. Okay, next term we have K is equal to 24 times five times five to the two minus one which is five. So we get four times five times five. And on the right hand side are second term is when J is one, we get four times five times five. Okay so these are equivalent but now we've written it to look like this general form and we can use our some a question. Let me just rewrite this once more. I'm just going to simplify this is 12 minus one. Okay just so we can rewrite it in this N minus one format so we can see what N. Is. And then we can multiply four by five is 20 and then we have five to the j. Okay so Our end value is 12 Are a value is 20. Our value is five. Okay we've compared to this general form the equation that we have and recall that the sum was given by a times one -R. to the end Divided by one -R. This is equal to 20 times one minus 5 to the exponent 12 Divided by -5. Okay and if you work this out on your calculator you're gonna get a great big number.  aN:aN:000NaN 1,220,703,120. Okay. And that is going to correspond with answer. D Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 31–36, find the indicated sum. Use the formula for the sum of the first n terms of a geometric sequence. 10 Σ (i = 1) 5 · 2^i

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