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JH
Numerade Educator

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Problem 50 Medium Difficulty

A sequence of terms is defined by
$ a_1 = 1 $ and $ a_n = (5 - n) a_{n-1} $
Calculate $ \sum_{n = 1}^{\infty} a_n $.

Answer

$\sum_{n=1}^{\infty} a_{n}=16$

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Video Transcript

a sequence is given by the following formula. The first term is one, so that's given. And then they tell you how to find all the remaining terms of the sequence. So a two will be five minus two and then we have a two minus one. So that's just three. A one, which is three times one here. So that's three. Let's find a few more terms. A three and then we have a three minus one. So it's two way too, and then a two is three. So two times three we'LL give you six a four. Only a few more terms here that will need and then ate to the four minus one. That is one times a three and we just found a three above That was six. And then now here this is where it gets interesting. Now we have five minus five and then a to the five minus one. So zero times a for but that she's going to be zero. And now what do we think? What a six will be? Five minus six a five. But we just saw a five zero. So this product is also zero, and we could see that for any end that's bigger than five A. N will be zero using the same reasoning because we'Ll always be multiplying by a zero for this term over here. So this means that we only need to add a few terms here. So this sum is just a one a two a three a four, a five and so on. But we just saw that everything starting at a five and after these are all zeros zero zero zero nce on. So we can ignore everything after an including a five because those do not contribute to the sun because you're adding zero. So we just need to add these first four out here. So we have won for the first one was one et tu. We found that to be three over here. Then we found a three and a four word, both six. So let's go ahead and add that together. So that's just ten here, plus six more and we get sixteen is our son, and that's the final answer

JH
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