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Find a formula for the general term $a_n$ of the sequence, assuming that the pattern of the first few terms continues.$\left\{\begin{array} 1, 0, -1, 0, 1, 0, -1, 0, . . . .\end{array}\right\}$

$1,0,-1,0,1,0,-1,0 \ldots .$

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Catherine R.

Missouri State University

Anna Marie V.

Campbell University

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Oregon State University

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University of Nottingham

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for this problem. It looks like there's not anything super nice happening. Looks like the odd terms. We're all going to be zero and the even terms they're going to be switching off between minus one at one. So it might be hard. Teo, write all of that information and one compact way, But you can always do it as this just writing it piece wise. So Anne is going to be zero if in is odd and it will be minus one to the in, divided by two if n is even right because we know that they even terms we're going to be switching between minus one and one. So if we just had minus one to the end happening here since when is even if we're looking at this bottom part than minus one to the end would always be a positive number. So that wouldn't be quite what we wanted, Since it is even we should be allowed to divide by two like we are here. You, Khun, we can check within his one. Then that means in his odds, we should get zero. So that agrees with what we have here. If it is too then we would have something. Even so, we'd be using this bottom part. We have minus one of the two over to which is just minus one. So that's good in equals. Three works in equals for Let's just check that in equals four. We would again use the this bottom part minus one to the four over too, would give us ah minus one squared, which is positive one, which is what we want. So this equation looks like it holds up.

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Catherine R.

Missouri State University

Anna Marie V.

Campbell University

Heather Z.

Oregon State University

Samuel H.

University of Nottingham

Lectures

Join Bootcamp