The nth term is Tn = (−1)(n−1)
Step-by-step explanation:
sequence = 1, -1, 1, -1, 1, -1, 1, -1, …. ?
This is a periodic, oscillating sequence. It is also something called a Geometric Progression where each term is a ratio. In your description you were expecting an arithmetic progression it seems- but it is not so.
For n=1
First term = 1
For n=2
Second term = -1
For n=3
Third term = 1
For n=4
Fourth term = -1
In every successive term the power is one less than the number of the particular term itself (for n=1th term the power was 0)
So we have this sequence:
(−1)0,(−1)1,(−1)2,(−1)3,(−1)4,(−1)5,…1,−1,1,−1,1,−1,…
To express this as a summation:
∑n=1∞(−1)(n−1)
So the nth term of the sequence in question will be (−1)(n−1) where n is the particular term of the sequence.
There is also a very simple formula you can use to arrive at the nth term of a geometric progression:
Tₙ = a⋅ (r (n−1) )
n is the nth term of the sequence
r is the common ratio
a is the first term of the sequence
So inserting these values:
Tn = 1⋅((−1)(n−1))
Tn = (−1)(n−1)