Question Video: Finding the Terms of a Sequence given Its General Term Mathematics

Find the first five terms of the sequence whose 𝑛th term is given by π‘Ž_(𝑛) = (βˆ’1)^(𝑛)/𝑛⁡, where 𝑛 β‰₯ 1.

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Video Transcript

Find the first five terms of the sequence whose 𝑛th term is given by π‘Ž sub 𝑛 equals negative one to the power of 𝑛 over 𝑛 to the fifth power, where 𝑛 is greater than or equal to one.

When we are given an 𝑛th term or a general term of a sequence, that allows us to work out any term value in the sequence. We can do this because the 𝑛th term will relate any terms in the sequence according to its position number or index 𝑛. Some sequences might start with an index 𝑛 equal to zero, but here we are told that 𝑛 must be greater than or equal to one. That means that the sequence will begin with an index of one. We can work out the first five terms in the sequence by substituting in different values of 𝑛. So into the 𝑛th term, we will substitute in the values of 𝑛 equals one, two, three, four, and five.

And so, for the first term, which will be π‘Ž sub one, we have negative one to the power of one over one to the power of five. On the numerator, negative one to the power of one is negative one. And on the denominator, one to the power of five is one. And of course, negative one over one is simply negative one. This means that the first term in the sequence will be negative one.

Let’s work out what the second term will be. Substituting in 𝑛 is equal to two into the 𝑛th term, we would have π‘Ž sub two is equal to negative one squared over two to the power of five. On the numerator, negative one squared will be one. And on the denominator, we recall that two to the power of five is 32. And so, the second term in the sequence is one over 32.

For the third term, we substitute 𝑛 equals three. And so, we have π‘Ž sub three is equal to negative one cubed over three to the power of five. That equals negative one over 243. We can evaluate the fourth and fifth terms in the same way to give one over 1024 and negative one over 3125. We might notice at this point that the terms in the sequence are alternating between positive and negative, which means it could be defined as an alternating sequence. This is caused by the term on the numerator in the 𝑛th term.

Observe that we are taking negative one to a power which is the index. And so every index which is even will create a positive term value. But here, we can give the answer that the first five terms in this sequence are negative one, one over 32, negative one over 243, one over 1024, and negative one over 3125.

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