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Exercise 8.1 · Q8

Q.Find the indicated term in the sequence whose nnth term is an=n22na_n = \dfrac{n^2}{2^n}; a7a_7.

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To find a7a_7, substitute n=7n=7 into the formula an=n22na_n = \frac{n^2}{2^n}. This gives a7=49128a_7 = \frac{49}{128}.

The core idea here is direct substitution: 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 number. No pattern-spotting, no recursion — just careful arithmetic.

Let's walk through it.

  1. Identify what's being asked.

    The sequence is defined by an=n22na_n = \dfrac{n^2}{2^n}. We need a7a_7, which means n=7n = 7.

  2. Substitute n=7n = 7 into the numerator.

    The numerator is n2n^2, so 72=497^2 = 49.

  3. Substitute n=7n = 7 into the denominator.

    The denominator is 2n2^n, so 27=1282^7 = 128.

    (If you need to compute powers of 2 quickly: 21=22^1=2, 22=42^2=4, 23=82^3=8, 24=162^4=16, 25=322^5=32, 26=642^6=64, 27=1282^7=128.)

  4. Write the fraction.

    So a7=49128a_7 = \dfrac{49}{128}.

  5. Check if simplification is possible.

    49=7249 = 7^2 and 128=27128 = 2^7 share no common factors (since 4949 is odd and 128128 is a power of 2), so the fraction is already in simplest form. …

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