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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.$$\left\{\begin{array} 4, -1, \frac {1}{4}, - \frac {1}{16}, \frac {1}{64}, . . . . .\end{array}\right\}$$

## $a_{n}=\frac{(-1)^{n}}{4^{n-1}}$

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we can see that we're going to be alternating signs or first term is negative. Next term is positive. Next term is negative. Next term is positive. So we should have some power of minus one happening. And the next thing we can observe is that there's some powers of four happening in the denominator. So you might gas that it's something like this. But then you can always do your check for any equals one. What we get, friend equals one would get minus one over four. So that's not what we want. Looks like we have the correct exponents for the over the minus one. But exponents for the four is not quite right. Case. Then you just have to modify it. All right, Well, what if we have minus one to the end? Divided by four to the n minus one. Hopefully that'LL give us the correct answer now and then you can check that minus one over four to zero, which is just minus one. So that looks good. And you can check for any equals two and equals two. You'd have a positive number One over four. That looks like it's good

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