Question

How to find and write the first four terms of the sequences below with the given general term? (1) `a_n=(-1)^(n+1)`(n+4) (2) `a_n=(-1)^(n-1)n^2`

Answer 1

For any sequence, the first term a1 is when n=1, the second term a2 is when n=2 and so on.
For your first sequence `a_1=(-1)^(1+1)(1+4)=5`
`a_2=(-1)^(2+1)(2+4)=-6`
continue for each term

Explanation:

Replace n by the whole number indicating the term you are calculating. A sequence is a list of terms in a specific order, the order is given by n, the value is given when you substitute n into the formula.

Answer 2

`(1) +5,-6,+7,-8`
`(2) +1,-4,+9,-16`

Explanation:

`(1)`
`a_n=(-1)^(n+1) (n+4)`

The first 4 terms of the sequence with general term `(n+4)` are: `5,6,7,8`

`(-1)^(n+1)` simply means that odd terms will be positive and even terms will be negative.

Hence, `a_(n=1->4)=(-1)^(n+1) (n+4) = +5,-6,+7,-8`

`(2)`
`a_n=(-1)^(n-1) n^2`

The first 4 terms of the sequence with general term `n^2` are: `1,4,9,16`

`(-1)^(n-1)` also means that odd terms will be positive and even terms will be negative. [Since, `(-1)^0 = 1`]

Hence, `a_(n=1->4)=(-1)^(n-1) n^2 = +1,-4,+9,-16`