Question Video: Finding the 𝑛th Term of an Unknown Sequence Mathematics

Consider the sequence 1, 1, 3/4, 4/8, …. Which of the following is the general term of this sequence such that 𝑛 β‰₯ 0? [A] 𝑛/2^𝑛 [B] (𝑛 βˆ’ 1)/2^𝑛 [C] (𝑛 + 1)/2^𝑛 [D] 2𝑛/2^𝑛 [E] (𝑛 + 2)/2^𝑛

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Video Transcript

Consider the sequence one, one, three-quarters, four-eighths, and so on. Which of the following is the general term of this sequence such that 𝑛 is greater than or equal to zero? Option (A) 𝑛 over two to the power of 𝑛. Option (B) 𝑛 minus one over two to the power of 𝑛. Option (C) 𝑛 plus one over two to the power of 𝑛. Option (D) two 𝑛 over two to the power of 𝑛. Or option (E) 𝑛 plus two over two to the power of 𝑛.

In this question, we’re given a sequence and asked to find its general term. When we’re finding a general term, we’re really finding a rule that connects the term number with the actual value of the term. When we’re given that the index 𝑛 is greater than or equal to zero, that means that our sequence begins with the zeroth term. We then have the first term π‘Ž sub one, the second term π‘Ž sub two, and so on. The 𝑛th term would be π‘Ž sub 𝑛. So given any value of 𝑛, what would the value in the sequence be? If we consider this sequence, there isn’t a common difference between any two consecutive terms, so this isn’t an arithmetic sequence. There also isn’t a common ratio between any two consecutive terms, so this isn’t a geometric sequence either.

In order to find the general term of the sequence, we’ll have to apply some logic. Let’s take a closer look at this term, π‘Ž sub one with the value of one. What if instead of being this value of one, this value of π‘Ž sub one was actually a fraction which simplified to one? In order for a fraction to simplify to one, the numerator and denominator would have to have the same value. Let’s say this fraction was actually two over something, and to simplify to one, it would need to be two over two. If we think of the zeroth term π‘Ž sub zero as instead of just being one as being a fraction of one over one, now we can see that the numerators actually have quite a nice pattern. They go from one to two to three to four. The denominators also have a different pattern. They go from one to two to four to eight.

Let’s consider the general term of the numerators and denominators separately for each value of 𝑛 starting with 𝑛 equals zero. Remember that we picked zero because this was given to us in the question. So for any index 𝑛, what will the numerator be? Well, every value in the numerator is one more than its index. So the 𝑛th term of the numerator will be 𝑛 plus one. For the denominators, these have a pattern which appears to be doubling. In fact, each denominator is a power of two. It would be two to the power of 𝑛. For example, for the zeroth term, two to the power of zero gives us one. For the first term, two to the power of one gives us two and so on. We can now put together the general term for the numerator and denominator. So the general term of this sequence is 𝑛 plus one over two to the power of 𝑛, which was the value given to us in option (C).

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