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leva [86]
3 years ago
9

The infinite sequence 1, 6, 15, 28, 45, ... exhibits which pattern? multiply the term number by 2, subtract 1 from the result, m

ultiply by the term number, and divide the result by 2. multiply the term number by 2, multiply the result by the term number, subtract 1, and divide the result by 2. multiply the term number by 2, subtract 1 from the result, multiply by 2 times the term number, and divide the result by 2. multiply the term number by 2, multiply the result by 2 times the term number, subtract 1, and divide the result by 2.
Mathematics
2 answers:
OlgaM077 [116]3 years ago
6 0
These are so-called  Hexagonal numbers. The general formula for the n-th hexagonal number is:
h_n=2n^2-n
h_n=\frac{2n(2n-1)}{2}
Let's compute a couple of them:
h_1=2(1)^2-1=1\\
h_2=2(2)^2-2=8-2=6\\
h_3=2(3)^2-3=18-3=15\\
h_4=2(4)^2-4=32-4=28\\
h_5=2(5)^2-5=50-5=45\\

Indeed this is our pattern.
The answer is:
multiply the term number by 2, subtract 1 from the result, multiply by 2 times the term number, and divide the result by 2.
Amanda [17]3 years ago
5 0
The correct response is ;
<span>multiply the term number by 2, subtract 1 from the result, multiply by 2 times the term number, and divide the result by 2.
if term number is x,
the number in the pattern for xth term is ;
multiply term number by 2 - 2x
subtract 1 - 2x - 1
multiply by 2 times the term number - (2x-1) * 2x
divide by 2 - (</span>(2x-1) * 2x)/2
so first term x=1
((2*1-1) * 2*1)/2
2/2 = 1
second term x = 2
((2*2-1) * 2*2)/2
(3*4)/2 = 6
third term x = 3
((2*3-1) * 2*3)/2
5*6/2 = 15
fourth term  x= 4
((2*4-1) * 2*4)/2
56/2 = 28
Therefore the above response shows the correct pattern
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