We read your photo as:

Find the term independent of p in the binomial expansion of ((3p2)/(2) - (1)/(3p))9.

Mathematics

ScanSolve AI Answer

Answer

D\text{D}

Step-by-step solution

8 steps

To find the term independent of pp in the binomial expansion of (3p22−13p)9\left(\frac{3p^2}{2} - \frac{1}{3p}\right)^9, we use the general term formula for a binomial expansion (a+b)n(a+b)^n, which is Tr+1=(nr)an−rbrT_{r+1} = \binom{n}{r} a^{n-r} b^r.

  1. Identify aa, bb, and nn.

    In this expansion, a=3p22a = \frac{3p^2}{2}, b=−13pb = -\frac{1}{3p}, and n=9n = 9.

  2. Write the general term Tr+1T_{r+1}.

    Tr+1=(9r)(3p22)9−r(−13p)rT_{r+1} = \binom{9}{r} \left(\frac{3p^2}{2}\right)^{9-r} \left(-\frac{1}{3p}\right)^r

  3. Separate the coefficients and the powers of pp.

    Tr+1=(9r)(32)9−r(p2)9−r(−13)r(p−1)rT_{r+1} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} (p^2)^{9-r} \left(-\frac{1}{3}\right)^r \left(p^{-1}\right)^r Tr+1=(9r)(32)9−r(−13)rp2(9−r)p−rT_{r+1} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r p^{2(9-r)} p^{-r} Tr+1=(9r)(32)9−r(−13)rp18−2r−rT_{r+1} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r p^{18-2r-r} Tr+1=(9r)(32)9−r(−13)rp18−3rT_{r+1} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r p^{18-3r}

  4. Find the value of rr for the term independent of pp.

    For the term to be independent of pp, the exponent of pp must be 0.

    18−3r=018 - 3r = 0 3r=183r = 18 r=6r = 6

  5. Substitute r=6r=6 back into the general term to find the constant term.

    The term independent of pp is T6+1=T7T_{6+1} = T_7.

    T7=(96)(32)9−6(−13)6T_7 = \binom{9}{6} \left(\frac{3}{2}\right)^{9-6} \left(-\frac{1}{3}\right)^6 T7=(96)(32)3(−13)6T_7 = \binom{9}{6} \left(\frac{3}{2}\right)^3 \left(-\frac{1}{3}\right)^6

  6. Calculate the binomial coefficient and the powers.

    (96)=(93)=9×8×73×2×1=3×4×7=84\binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84 (32)3=3323=278\left(\frac{3}{2}\right)^3 = \frac{3^3}{2^3} = \frac{27}{8} (−13)6=(−1)636=1729\left(-\frac{1}{3}\right)^6 = \frac{(-1)^6}{3^6} = \frac{1}{729}

  7. Multiply the calculated values.

    T7=84×278×1729T_7 = 84 \times \frac{27}{8} \times \frac{1}{729} T7=84×278×729T_7 = \frac{84 \times 27}{8 \times 729}

    Simplify the expression:

    T7=848×27729T_7 = \frac{84}{8} \times \frac{27}{729} T7=212×127T_7 = \frac{21}{2} \times \frac{1}{27} T7=2154T_7 = \frac{21}{54}

    Divide both numerator and denominator by their greatest common divisor, which is 3:

    T7=21÷354÷3=718T_7 = \frac{21 \div 3}{54 \div 3} = \frac{7}{18}

  8. Compare the result with the given options.

    The calculated term is 718\frac{7}{18}, which corresponds to option D.

    The final answer is D\text{D}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QFind the binomial expansion of x2(x−2y)3 and the term independent of x in the expansion of (x…QFind the term independent of x in the expansion of (x − (3)/(x))4.QFind the binomial expansion of x2(x−2y)3. 2. Find the term independent of x in the expansion of (x…QIdentify the general term in the binomial expansion.QFind the coefficient of x4 in the expansion of ((1)/(p) + px)7.
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp