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Find the 6 th term in the expansion of (4x-1)^(7) in simplest form. Answer:

Find the 66 th term in the expansion of (4x−1)7 (4 x-1)^{7} in simplest form.\newlineAnswer:

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Full solution

Q. Find the 66 th term in the expansion of (4x−1)7 (4 x-1)^{7} in simplest form.\newlineAnswer:
  1. Use Binomial Theorem: To find the 6th6^{\text{th}} term in the expansion of (4x−1)7(4x-1)^{7}, we will use the binomial theorem. The general term in the expansion of (a+b)n(a+b)^{n} is given by T(k+1)=(nk)⋅an−k⋅bkT(k+1) = \binom{n}{k} \cdot a^{n-k} \cdot b^{k}, where (nk)\binom{n}{k} is the binomial coefficient ＂n choose k＂. For the 6th6^{\text{th}} term, k=5k = 5, since we start counting from k=0k = 0 for the first term.
  2. Calculate Binomial Coefficient: First, we calculate the binomial coefficient for n=7n = 7 and k=5k = 5, which is 7C57C5. This can be calculated as 7!5!×(7−5)!\frac{7!}{5! \times (7-5)!}.
  3. Find 7C57C5: Calculating 7C57C5, we get 7!5!⋅2!=7⋅62⋅1=21\frac{7!}{5! \cdot 2!} = \frac{7 \cdot 6}{2 \cdot 1} = 21.
  4. Calculate 66th Term: Now, we will use the binomial coefficient to find the 66th term. The 66th term is T(6)=(75)⋅(4x)7−5⋅(−1)5T(6) = \binom{7}{5} \cdot (4x)^{7-5} \cdot (-1)^5.
  5. Substitute Values: Substitute the values into the formula to get T(6)=21×(4x)2×(−1)5T(6) = 21 \times (4x)^2 \times (-1)^5.
  6. Simplify Term: Simplify the term: T(6)=21×16x2×(−1)T(6) = 21 \times 16x^2 \times (-1).
  7. Final Result: Multiplying the numbers together, we get T(6)=−336x2T(6) = -336x^2.

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