The nth term is 2n2.
(One way to find that is to notice at all the numbers are even, then divide them by 2. The sequence becomes 1, 4, 9, 16, 25, which are the square numbers in order.)
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To find the nth term of a sequence, we first need to determine the pattern or rule governing the sequence. In this case, the sequence appears to be increasing by consecutive odd numbers (2, 4, 6, 8, etc.). The nth term of this sequence can be represented by the formula n^2 + 1. So, the nth term of the sequence 2, 8, 18, 32, 50 is n^2 + 1.
# in python
n = 10
for i in range(n):
print(i, i**2*2)
'''
0 0
1 2
2 8
3 18
4 32
5 50
6 72
7 98
8 128
9 162
'''
The answer is 128/(2^(n-1)) if the 1st term is 128. The divisor is found by the realization that these are decreasing powers of two.
29, assuming it is an algebraically reclusive sequence.
Well, darling, the first 5 terms in that fancy sequence are 28, 26, 24, 22, and 20. You get those numbers by plugging in n values 1 through 5 into the formula 30-2n. So, there you have it, sweet cheeks!
This sequence is called the doubling sequence.
32 + 25x + 18 = 50 + 25x