### Video Transcript

Which of the following represents the general term of the sequence two, negative five, 10, negative 17, and so on. Option (A) negative one to the power of π times π squared plus one. Option (B) negative one to the power of π times π squared minus one. Option (C) negative one to the power of π plus one times π squared plus one. Option (D) negative one to the power of two π times π squared plus one. Or option (E) negative one to the power of π plus one times π squared minus one.

Each of the answer options here represents different options for the possible general term of this sequence of values. The general term or the πth term relates values of the index π to the actual term value. If we assume that the values of the index π are integers which are greater than or equal to one, then that means that the first term in the sequence would have an index of one. The second term would have an index of two and so on.

We can use the answer options here to help us work out what the general term will be. For example, if we substitute in the values of π equals one, two, three, four in turn into any of these general terms and it gave us the values of two, negative five, 10, and negative 17, respectively, then that would mean that it is the general term of this sequence. So letβs start by substituting π equals one into each of the general terms that we were given.

So for the πth term in option (A), when π is equal to one, we would have negative one to the power of one times one squared plus one. We know that negative one to the power of one will be negative one. And when we multiply that by two, we would get negative two. In option (B), the first term π sub one can be given as negative one to the power of one times one squared minus one. We would simplify this to get negative one to the power of one. Thatβs negative one. But the second set of parentheses will evaluate as zero, and negative one times zero is zero.

The first term in option (C) can be given as negative one to the power of one plus one times one squared plus one. This time, we will have negative one to the power of two, which means that the first set of parentheses here would simplify to one. And one times two is two. In the same way, we can evaluate the first term of the general term given in option (D) as two and in option (E) as zero. Remember that the first term in the sequence that we were given is two. There are only two general terms here which give a first term of two. And so we can eliminate the general terms in options (A), (B), and (E).

We will now substitute π equals two into each of the general terms given in options (C) and (D). For option (C) then, when π is equal to two, we have negative one to the power of two plus one times two squared plus one. We can simplify this to negative one cubed times four plus one. Working this out, we have negative one times five, which is negative five. Remember that weβre looking for the second term in the sequence to be negative five.

So far, this general term looks good, but letβs check the general term given in option (D). Substituting π equals two, we have negative one to the power of two times two multiplied by two squared plus one. So we have negative one to the power of four times four plus one. This simplifies to one times five, which gives us a second term of five. But remember that weβre looking for a second term of negative five. And so that means that the second term of this general term indeed does not fit. Therefore, we can eliminate option (D).

As an aside, if we look at the πth term given in option (D), the fact that we have negative one to the power of two π means that the power of negative one will always be even. And so any term produced by this general term given in option (D) will always be positive. The sequence of values that we were given is not always positive. So option (D) will definitely not work.

We have now eliminated four out of the five answer options. But we should check that option (C) also produces the same third and forth terms that we have in the sequence.

Substituting π equals three into the general term negative one to the power of π plus one times π squared plus one gives us negative one to the power of three plus one times three squared plus one, which will evaluate as 10. To find the fourth term, we substitute π equals four into the general term. And that gives us an answer of negative 17. And so the general term in option (C) gave us a first term of two, a second term of negative five, a third term of 10, and a fourth term of negative 17.

We can then give the answer that the general term of the sequence two, negative five, 10, and negative 17 is represented by that given in option (C). Itβs negative one to the power of π plus one times π squared plus one.