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Exercise 8.1 · Q10

Q.Find the indicated term in the sequence whose nnth term is an=n(n−2)n+3a_n = \dfrac{n(n-2)}{n+3}; a20a_{20}.

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To find a20a_{20}, substitute n=20n = 20 directly into the formula an=n(n−2)n+3a_n = \frac{n(n-2)}{n+3} and simplify. The result is a20=36023a_{20} = \frac{360}{23}.

The core idea here is Sequence Term Evaluation: when you're given an explicit formula for the nnth term of a sequence, finding any specific term is just a matter of plugging in the index. No recursion, no pattern-spotting — just careful substitution and arithmetic.

Let's walk through it.

  1. Identify what's being asked.

    The nnth term is an=n(n−2)n+3a_n = \dfrac{n(n-2)}{n+3}. We need a20a_{20}, which means n=20n = 20.

  2. Substitute n=20n = 20 into the formula.

    Replace every nn with 2020:

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

  1. Simplify inside the parentheses first. 20−2=1820 - 2 = 18, and 20+3=2320 + 3 = 23. So:

a20=20×1823a_{20} = \frac{20 \times 18}{23}

  1. Multiply the numerator. 20×18=36020 \times 18 = 360. So:

a20=36023a_{20} = \frac{360}{23}

  1. Check if simplification is possible. 360360 and 2323 share no common factors (23 is prime and does not divide 360), so the fraction is already in simplest form. …

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