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Blaine Woodward
Mathematics
17 March, 04:17
Find the nth tern of the sequence 1,1,1,1
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America Griffith
17 March, 04:42
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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)
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