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

Find the first five terms of the sequence whose πth term is given by π_π = (β1)^π/πβ΅, where π β₯ 1.

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### Video Transcript

Find the first five terms of the sequence whose πth term is given by π sub π is equal to negative one to the power of π divided by π to the fifth power, where π is greater than or equal to one.

In order to work out the first five terms of a sequence where π is greater than or equal to one, we need to substitute the numbers one, two, three, four, and five into the πth-term formula. This will give us values for π sub one through π sub five. When π is equal to one, we have negative one to the power of one divided by one to the fifth power. When π is equal to two, we have negative one squared over two to the fifth power. When π equals three, we have negative one cubed over three to the fifth power. The fourth and fifth terms, π four and π five, are as shown.

When raising negative one to an odd power, our answer will be negative one. This means that the numerator of our first, third, and fifth terms will be negative one. Negative one raised to an even power will give us positive one. This means that the numerator for our second and fourth term will be positive. One to the fifth power is just equal to one. Two to the fifth power is 32. Three to the fifth power is 243, four to the fifth power 1024. And five to the fifth power is 3125. Negative one divided by one is just negative one. So, π sub one is negative one.

None of the other four fractions can be simplified. Therefore, the first five terms of the sequence are negative one, one over 32, negative one over 243, one over 1024, and negative one over 3125.