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The general term in the binomial expansion of (a+b)ⁿ is given by the formula:

Mathematics

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B

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The general term in the binomial expansion of (a+b)n(a+b)^n is given by the formula:

Tk+1=(nk)an−kbkT_{k+1} = \binom{n}{k} a^{n-k} b^k

where (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}.

: Obtain the fourth term in the expansion of (x−y)9(x-y)^9.
Here, a=xa=x, b=−yb=-y, n=9n=9. For the fourth term, k+1=4k+1=4, so k=3k=3.

  1. Apply the general term formula.

    T4=(93)(x)9−3(−y)3T_4 = \binom{9}{3} (x)^{9-3} (-y)^3
  2. Calculate the binomial coefficient.

    (93)=9!3!(9−3)!=9!3!6!=9×8×73×2×1=3×4×7=84\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84
  3. Substitute the values and simplify.

    T4=84x6(−y3)T_4 = 84 x^6 (-y^3) T4=−84x6y3T_4 = -84x^6y^3

    The correct option is B.
    The fourth term is B. −84x6y3\boxed{\text{B. } -84x^6y^3}.

    : Obtain the fourth term in the expansion of (x2+3)6(x^2 + 3)^6.
    Here, a=x2a=x^2, b=3b=3, n=6n=6. For the fourth term, k+1=4k+1=4, so k=3k=3.

  4. Apply the general term formula.

    T4=(63)(x2)6−3(3)3T_4 = \binom{6}{3} (x^2)^{6-3} (3)^3
  5. Calculate the binomial coefficient.

    (63)=6!3!(6−3)!=6!3!3!=6×5×43×2×1=20\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
  6. Substitute the values and simplify.

    T4=20(x2)3(33)T_4 = 20 (x^2)^3 (3^3) T4=20(x6)(27)T_4 = 20 (x^6) (27) T4=540x6T_4 = 540x^6

    The correct option is B.
    The fourth term is B. 540x6\boxed{\text{B. } 540x^6}.

    : Obtain the sixth term in the expansion of (3x−2y)9(3x - 2y)^9.
    Here, a=3xa=3x, b=−2yb=-2y, n=9n=9. For the sixth term, k+1=6k+1=6, so k=5k=5.

  7. Apply the general term formula.

    T6=(95)(3x)9−5(−2y)5T_6 = \binom{9}{5} (3x)^{9-5} (-2y)^5
  8. Calculate the binomial coefficient.

    (95)=9!5!(9−5)!=9!5!4!=9×8×7×64×3×2×1=9×2×7=126\binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 9 \times 2 \times 7 = 126
  9. Substitute the values and simplify.

    T6=126(3x)4(−2y)5T_6 = 126 (3x)^4 (-2y)^5 T6=126(34x4)((−2)5y5)T_6 = 126 (3^4 x^4) ((-2)^5 y^5) T6=126(81x4)(−32y5)T_6 = 126 (81 x^4) (-32 y^5) T6=126×81×(−32)x4y5T_6 = 126 \times 81 \times (-32) x^4 y^5 T6=10206×(−32)x4y5T_6 = 10206 \times (-32) x^4 y^5 T6=−326592x4y5T_6 = -326592 x^4 y^5

    The correct option is D.
    The sixth term is D. −326592x4y5\boxed{\text{D. } -326592x^4y^5}.

    : Obtain the second term in the expansion of (2x−1)2(2x - 1)^2.
    Here, a=2xa=2x, b=−1b=-1, n=2n=2. For the second term, k+1=2k+1=2, so k=1k=1.

  10. Apply the general term formula.

    T2=(21)(2x)2−1(−1)1T_2 = \binom{2}{1} (2x)^{2-1} (-1)^1
  11. Calculate the binomial coefficient.

    (21)=2!1!(2−1)!=2!1!1!=2\binom{2}{1} = \frac{2!}{1!(2-1)!} = \frac{2!}{1!1!} = 2
  12. Substitute the values and simplify.

    T2=2(2x)1(−1)1T_2 = 2 (2x)^1 (-1)^1 T2=2(2x)(−1)T_2 = 2 (2x) (-1) T2=−4xT_2 = -4x

    The correct option is C.
    The second term is C. −4x\boxed{\text{C. } -4x}.

    : What is the coefficient of x3x^3 in (2+3x)4(2 + 3x)^4?
    Here, a=2a=2, b=3xb=3x, n=4n=4. We need to find the term where the power of xx is 3.
    The general term is Tk+1=(nk)an−kbkT_{k+1} = \binom{n}{k} a^{n-k} b^k.

  13. Write the general term for this expansion.

    Tk+1=(4k)(2)4−k(3x)kT_{k+1} = \binom{4}{k} (2)^{4-k} (3x)^k Tk+1=(4k)(2)4−k(3kxk)T_{k+1} = \binom{4}{k} (2)^{4-k} (3^k x^k)
  14. Identify the value of kk for x3x^3.

    For the term containing x3x^3, we must have xk=x3x^k = x^3, so k=3k=3.

  15. Substitute k=3k=3 into the general term formula.

    T3+1=T4=(43)(2)4−3(33x3)T_{3+1} = T_4 = \binom{4}{3} (2)^{4-3} (3^3 x^3) T4=(43)(2)1(33)x3T_4 = \binom{4}{3} (2)^1 (3^3) x^3
  16. Calculate the binomial coefficient and simplify.

    (43)=4!3!(4−3)!=4!3!1!=4\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = 4 T4=4×2×27x3T_4 = 4 \times 2 \times 27 x^3 T4=8×27x3T_4 = 8 \times 27 x^3 T4=216x3T_4 = 216 x^3

    The coefficient of x3x^3 is 216.
    The correct option is D.
    The coefficient of x3x^3 is D. 216\boxed{\text{D. } 216}.

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