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Numerade Educator



Problem 17 Easy Difficulty

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} \frac {1}{2}, - \frac {4}{3}, \frac {9}{4}, - \frac {16}{5}, \frac {25}{6}, . . . . .\end{array}\right\} $


$a_{n}=\frac{(-1)^{n-1} n^{2}}{n+1}$


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Video Transcript

that first term supposed to be one half can't one thing that we notices that there's in the denominator all devising the denominator, always just increasing by one. They're always going to be increasing by one. And we need to add one here so that we start on, too. We should start on too. Okay, So if we have in plus one in the denominator, that means that we'LL start on too. But we'LL be one more. Every time I write this in, he's going to make us be one more in the previous term, every time in the denominator, and we're also going to be alternating signs. First term is positive. Next term is negative. Next term is positive. It's where all training signs so well, maybe something like this. Well, then you can check to see if you plug in and equals one. Then you'd have minus one to the one, which would give you a negative number. So then you need to modify that. If you put in minus one there, then you should start on a positive number, which is what you want, Kane. And the only other thing that we need is that we have powers and the numerator. So the first term we have one squared. Next term we have ah, four, which is two squared. The next term we have a nine, which is three squared. Next term we have a sixteen, which is four squared. So we should have in in squared here. Okay, so a n should be minus one to the end, minus one times and squared, divided by N plus one.