How do you Find the #n#-th term of the infinite sequence #1,-2/3,4/9,-8/27,…#?
When looking at sequences, your mind should always be in a pattern-recognizing mode. Sometimes sequences seem daunting, but when you look at them in pieces, they're usually easier to tackle.
With this sequence, we can see that for each term, the numerator doubles, and the denominator triples. Also, it's important to note that each successive term flips sign. (positive to negative to positive)
Now, we simply piece them together:
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The given sequence alternates between positive and negative terms and follows a pattern of powers of 2 in the numerators and powers of 3 in the denominators.
The nth term of the sequence can be expressed as follows:
[ a_n = (-1)^{n+1} \times \frac{2^{n-1}}{3^n} ]
where ( (-1)^{n+1} ) alternates the sign of the terms, ( 2^{n-1} ) represents the power of 2 in the numerator, and ( 3^n ) represents the power of 3 in the denominator.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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