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Worked Examples · Example 2

Q.What is the 20th term of the sequence defined by an=(n−1)(2−n)(3+n)a_n = (n-1)(2-n)(3+n)?

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The 20th term is found by substituting n=20n=20 into the given formula and simplifying. The result is −7866\boxed{-7866}.

The question gives us a direct formula for the nnth term: an=(n−1)(2−n)(3+n)a_n = (n-1)(2-n)(3+n). This is an explicit definition, not a recursive one — so finding any term is just a matter of plugging in the right value of nn.

The cleanest path is to substitute first, then simplify step by step, rather than expanding the product algebraically first.

  1. Substitute n=20n=20 into the formula:

a20=(20−1)(2−20)(3+20)a_{20} = (20-1)(2-20)(3+20)

  1. Simplify each bracket:

    • 20−1=1920-1 = 19
    • 2−20=−182-20 = -18
    • 3+20=233+20 = 23

    So:

a20=19×(−18)×23a_{20} = 19 \times (-18) \times 23

  1. Multiply step by step:
    • 19×(−18)=−34219 \times (-18) = -342
    • (−342)×23(-342) \times 23:
      • 342×20=6840342 \times 20 = 6840
      • 342×3=1026342 \times 3 = 1026
      • 6840+1026=78666840 + 1026 = 7866
    • So (−342)×23=−7866(-342) \times 23 = -7866 …

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