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Find the 
5^("th ") term in the expansion of 
(5x-y)^(6) in simplest form.
Answer:

Find the 5th  5^{\text {th }} term in the expansion of (5xy)6 (5 x-y)^{6} in simplest form.\newlineAnswer:

Full solution

Q. Find the 5th  5^{\text {th }} term in the expansion of (5xy)6 (5 x-y)^{6} in simplest form.\newlineAnswer:
  1. Identify general term: Identify the general term of a binomial expansion.\newlineThe general term in the expansion of (ab)n(a - b)^n is given by T(k+1)=C(n,k)a(nk)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 5th5^{\text{th}} term, which corresponds to k=4k = 4, since the first term corresponds to k=0k = 0.
  3. Apply binomial theorem: Apply the binomial theorem to find the 5th5^{\text{th}} term. Using the formula from Step 11, we substitute n=6n = 6 and k=4k = 4 to find T(5)=C(6,4)(5x)64(y)4T(5) = C(6, 4) \cdot (5x)^{6-4} \cdot (-y)^4.
  4. Calculate binomial coefficient: Calculate the binomial coefficient C(6,4)C(6, 4).C(6,4)=6!4!(64)!=(654321)((4321)(21))=(65)(21)=15C(6, 4) = \frac{6!}{4! \cdot (6-4)!} = \frac{(6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{((4 \cdot 3 \cdot 2 \cdot 1) \cdot (2 \cdot 1))} = \frac{(6 \cdot 5)}{(2 \cdot 1)} = 15.
  5. Substitute values: Substitute the values into the term.\newlineT(5)=15×(5x)2×(y)4=15×25x2×y4T(5) = 15 \times (5x)^2 \times (-y)^4 = 15 \times 25x^2 \times y^4.
  6. Simplify term: Simplify the term. T(5)=15×25×x2×y4=375×x2×y4T(5) = 15 \times 25 \times x^2 \times y^4 = 375 \times x^2 \times y^4.

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