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.
Oh, what a lovely sequence we have here! To find the first 5 terms, we simply substitute n with 1, 2, 3, 4, and 5 into the formula 30-2n. So, the first 5 terms are 28, 26, 24, 22, and 20. Just like painting a beautiful landscape, each term in this sequence has its own unique beauty.
This sequence is called the doubling sequence.
32 + 25x + 18 = 50 + 25x