Give that coefficient of x³ in the expansion of [a+x³ \2 ]⁴ is 1.Find the value of a

Mathematics

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Answer

2

Step-by-step solution

6 steps
  1. Write the general term for the binomial expansion of (a+x2)4\left(a + \frac{x}{2}\right)^4.

    The general term Tr+1T_{r+1} is given by:

    Tr+1=(nr)An−rBrT_{r+1} = \binom{n}{r} A^{n-r} B^r

    In this case, A=aA=a, B=x2B=\frac{x}{2}, and n=4n=4.

    Tr+1=(4r)a4−r(x2)rT_{r+1} = \binom{4}{r} a^{4-r} \left(\frac{x}{2}\right)^r Tr+1=(4r)a4−rxr2rT_{r+1} = \binom{4}{r} a^{4-r} \frac{x^r}{2^r}
  2. Find the term containing x3x^3.

    For the term containing x3x^3, we need r=3r=3.
    Substitute r=3r=3 into the general term:

    T3+1=(43)a4−3x323T_{3+1} = \binom{4}{3} a^{4-3} \frac{x^3}{2^3} T4=(43)a1x38T_4 = \binom{4}{3} a^1 \frac{x^3}{8}
  3. Calculate the binomial coefficient (43)\binom{4}{3}.

    (43)=4!3!(4−3)!=4!3!1!=4×3×2×1(3×2×1)(1)=4\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(1)} = 4
  4. Substitute the value of (43)\binom{4}{3} back into the term.

    T4=4⋅a⋅x38T_4 = 4 \cdot a \cdot \frac{x^3}{8} T4=4a8x3T_4 = \frac{4a}{8} x^3 T4=a2x3T_4 = \frac{a}{2} x^3
  5. Identify the coefficient of x3x^3 and set it equal to 1.

    The coefficient of x3x^3 is a2\frac{a}{2}.
    Given that the coefficient of x3x^3 is 1:

    a2=1\frac{a}{2} = 1
  6. Solve for aa.

    a=1×2a = 1 \times 2 a=2a = 2

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

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