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Find the 6 th term in the expansion of (x-3y)^(7) in simplest form. Answer:

Find the 66 th term in the expansion of (x−3y)7 (x-3 y)^{7} in simplest form.\newlineAnswer:

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Full solution

Q. Find the 66 th term in the expansion of (x−3y)7 (x-3 y)^{7} in simplest form.\newlineAnswer:
  1. Identify general term: Identify the general term for the binomial expansion.\newlineThe general term in the expansion of (a−b)n(a - b)^n is given by T(k+1)=C(n,k)⋅a(n−k)⋅bkT(k+1) = C(n, k) \cdot a^{(n-k)} \cdot b^k, where C(n,k)C(n, k) is the binomial coefficient ＂n choose k＂.
  2. Determine specific term: Determine the specific term we are looking for.\newlineWe want to find the 6th6^{\text{th}} term, which means k=5k = 5 because the first term corresponds to k=0k = 0.
  3. Calculate binomial coefficient: Calculate the binomial coefficient for the 66th term.\newlineC(7,5)=7!5!×(7−5)!=7×62×1=21C(7, 5) = \frac{7!}{5! \times (7-5)!} = \frac{7 \times 6}{2 \times 1} = 21.
  4. Substitute values: Substitute the values into the general term formula. T(6)=C(7,5)×x(7−5)×(−3y)5=21×x2×(−3y)5T(6) = C(7, 5) \times x^{(7-5)} \times (-3y)^5 = 21 \times x^2 \times (-3y)^5.
  5. Simplify term: Simplify the term. T(6)=21×x2×(−243y5)=−5103x2y5T(6) = 21 \times x^2 \times (-243y^5) = -5103x^2y^5.

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