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20 February, 02:45

How can the next term in the infinite sequence 1, 5, 12, 22, 35, ... be generated? Square the term number, subtract the term number from the result, multiply by 3, and divide the result by 2. Square the term number, multiply the result by 3, divide by 2, and subtract the term number from the result. Square the term number, divide the result by 2, subtract the term number, and multiply the result by 3. Square the term number, multiply the result by 3, subtract the term number, and divide the result by 2.

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  1. 20 February, 03:08
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    Hello,

    "Square the term number, multiply the result by 3, subtract the term number, and divide the result by 2" is the correct answer.

    Assume n the term number

    U (n) = (x²*3-x) / 2

    U (1) = (3*1-1) / 2=2/2=1

    U (2) = (3*4-2) / 2=10/2=5

    U (3) = (3*9-3) / 2=24/2=12

    U (4) = (3*16-4) / 2=44/2=22

    U (5) = (3*25-5) / 2=70/2=35

    U (6) = (3*36-6) / 2=102/2=51
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