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Identify the parameters of the binomial expansion.

Mathematics

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Answer

±2\pm 2

Step-by-step solution

6 steps
  1. Identify the parameters of the binomial expansion.

    The given expansion is (a2+2)8\left(\frac{a}{2} + 2\right)^8.
    Here, n=8n=8.
    The number of terms in the expansion is n+1=8+1=9n+1 = 8+1 = 9.

  2. Determine the position of the middle term.

    Since there are 9 terms, the middle term is the 9+12=5th\frac{9+1}{2} = 5^{th} term.
    For the general term Tr+1T_{r+1}, we have r+1=5r+1 = 5, so r=4r=4.

  3. Write the formula for the general term and substitute the values.

    The general term in the expansion of (x+y)n(x+y)^n is given by Tr+1=(nr)xn−ryrT_{r+1} = \binom{n}{r} x^{n-r} y^r.
    In this case, x=a2x = \frac{a}{2}, y=2y = 2, n=8n=8, and r=4r=4.

    T5=(84)(a2)8−4(2)4T_5 = \binom{8}{4} \left(\frac{a}{2}\right)^{8-4} (2)^4 T5=(84)(a2)4(2)4T_5 = \binom{8}{4} \left(\frac{a}{2}\right)^4 (2)^4 T5=(84)a424⋅24T_5 = \binom{8}{4} \frac{a^4}{2^4} \cdot 2^4 T5=(84)a4T_5 = \binom{8}{4} a^4
  4. Calculate the binomial coefficient (84)\binom{8}{4}.

    (84)=8!4!(8−4)!=8!4!4!=8×7×6×54×3×2×1\binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} (84)=168024\binom{8}{4} = \frac{1680}{24} (84)=70\binom{8}{4} = 70
  5. Set up the equation using the given value of the middle term.

    We are given that the middle term is 1120.

    70a4=112070 a^4 = 1120
  6. Solve for aa.

    a4=112070a^4 = \frac{1120}{70} a4=16a^4 = 16

    Take the fourth root of both sides:

    a=±164a = \pm \sqrt[4]{16} a=±2a = \pm 2

    The value of aa is ±2\boxed{\pm 2}.

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