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Find the 4 th term in the expansion of (x-3)^(8) in simplest form. Answer:

Find the 44 th term in the expansion of (x−3)8 (x-3)^{8} in simplest form.\newlineAnswer:

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Full solution

Q. Find the 44 th term in the expansion of (x−3)8 (x-3)^{8} in simplest form.\newlineAnswer:
  1. Use Binomial Theorem: To find the 44th term in the expansion of (x−3)8(x-3)^{8}, we will use the binomial theorem. The binomial theorem states that the kkth term in the expansion of (a+b)n(a+b)^{n} is given by C(n,k−1)⋅a(n−k+1)⋅b(k−1)C(n, k-1) \cdot a^{(n-k+1)} \cdot b^{(k-1)}, where C(n,k)C(n, k) is the binomial coefficient ＂nn choose kk＂. For the 44th term, k=4k = 4.
  2. Calculate Binomial Coefficient: First, we calculate the binomial coefficient C(8,4−1)C(8, 4-1) which is C(8,3)C(8, 3). This is equal to 8!3!×(8−3)!\frac{8!}{3! \times (8-3)!}, where “!“\text{“!“} denotes factorial.
  3. Simplify Factorials: Calculating the factorial values, we get 8!=8×7×6×5×4×3×2×18! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1, 3!=3×2×13! = 3 \times 2 \times 1, and (8−3)!=5!=5×4×3×2×1(8-3)! = 5! = 5 \times 4 \times 3 \times 2 \times 1.
  4. Calculate Binomial Coefficient: Now, we simplify the binomial coefficient C(8,3)=8!3!×5!=8×7×63×2×1=56C(8, 3) = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56.
  5. Calculate Powers of a and b: Next, we calculate the remaining parts of the term: a(n−k+1)=x(8−4+1)=x5a^{(n-k+1)} = x^{(8-4+1)} = x^5 and b(k−1)=(−3)(4−1)=(−3)3b^{(k-1)} = (-3)^{(4-1)} = (-3)^3.
  6. Simplify Powers: Now, we simplify (−3)3=−27(-3)^3 = -27.
  7. Multiply Coefficients: Finally, we multiply the binomial coefficient by the powers of aa and bb to get the 44th term: 56×x5×(−27)56 \times x^5 \times (-27).
  8. Final Result: Multiplying 5656 by −27-27, we get −1512-1512. So, the 44th term in the expansion is −1512×x5-1512 \times x^5.

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