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

The 𝑛th term in a sequence is given by π‘Ž_(𝑛 + 1) = π‘›π‘Ž_(𝑛). Find the first six terms of this sequence, given that π‘Žβ‚ = βˆ’118.


Video Transcript

The 𝑛th term of a sequence is given by π‘Ž sub 𝑛 plus one equals π‘›π‘Ž sub 𝑛. Find the first six terms of this sequence, given that π‘Ž sub one equals negative 118.

We’ve been given a formula for finding the 𝑛th term or general term of a sequence. It’s π‘Ž sub 𝑛 plus one equals 𝑛 multiplied by π‘Ž sub 𝑛. Now, this is an example of a sequence being defined recursively, which means that each term is given as a function of the terms that come before it. What this recursive definition is telling us then is that each time, in order to find the next term in the sequence, we have to multiply the previous term by its own term number. So, let’s go ahead then and find the first six terms in this sequence, which we can label as π‘Ž sub one, π‘Ž sub two, all the way up to π‘Ž sub six.

We’re told in the question that the first term π‘Ž sub one is equal to negative 118. So, we can fill this value in directly. Now, let’s consider the second term in the sequence, which is π‘Ž sub two. Now, remember, our recursive definition uses π‘Ž sub 𝑛 plus one. So, π‘Ž sub two would be equivalent to π‘Ž sub one plus one, which means the value of 𝑛 that we’re using here is one. It isn’t actually the value of the term number for this term; it’s the term number for the previous term. So, substituting a value of 𝑛 equal to one into our recursive definition, we have that π‘Ž sub two will be equal to one multiplied by π‘Ž sub one. That’s one multiplied by negative 118 which is negative 118. So, we find that the second term in our sequence is actually the same as the first.

Now, we consider the third term, π‘Ž sub three, which we can think of as π‘Ž sub two plus one, meaning that the value of 𝑛 we’re going to use here is two. In our recursive definition then, we have that π‘Ž sub three is equal to two multiplied by π‘Ž sub two. That’s two multiplied by the previous term negative 118, which is negative 236. So, we found the third term in our sequence. In the same way then, we can find expressions for the fourth, fifth, and sixth terms in terms of their proceeding terms. π‘Ž sub four is equal to three multiplied by π‘Ž sub three. π‘Ž sub five is equal to four multiplied by π‘Ž sub four. And π‘Ž sub six is equal to five multiplied by π‘Ž sub five. This gives three multiplied by negative 236 which is negative 708 for the fourth term. Multiplying this by four gives negative 2832 for the fifth term. And finally, multiplying this value by five gives negative 14160 for the sixth term.

Remember what we’ve done each time to create this sequence from its recursive definition is multiplying the previous term by the previous term number. So, we multiply by one, then two, then three, then four, and then five.

So, we found that the first six terms of this sequence are negative 118, negative 118, negative 236, negative 708, negative 2832, and negative 14160.

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