Find the nth tern of the sequence 1,1,1,1

The nth term is Tn = (−1)(n−1)

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)

b) well because you get well

typically if you want , you would actually put questions here for us to you. but hey, that's just me. maybe someone will magically answer the questions for you without knowing what they are.