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Find the 2 nd term in the expansion of (4x+7y)^(5) in simplest form. Answer:

Find the 22 nd term in the expansion of (4x+7y)5 (4 x+7 y)^{5} in simplest form.\newlineAnswer:

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Full solution

Q. Find the 22 nd term in the expansion of (4x+7y)5 (4 x+7 y)^{5} in simplest form.\newlineAnswer:
  1. Use Binomial Theorem: To find the 22nd term in the expansion of (4x+7y)5(4x+7y)^{5}, 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)=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＂. For the 22nd term, k=1k=1.
  2. Calculate Binomial Coefficient: Calculate the binomial coefficient for the 22nd term, which is C(5,1)C(5, 1). C(5,1)=5!1!∗(5−1)!=51=5C(5, 1) = \frac{5!}{1! * (5-1)!} = \frac{5}{1} = 5.
  3. Apply General Term Formula: Now, plug in the values into the general term formula for k=1k=1. T(2)=C(5,1)×(4x)(5−1)×(7y)1T(2) = C(5, 1) \times (4x)^{(5-1)} \times (7y)^1.
  4. Simplify Expression: Simplify the expression. T(2)=5×(4x)4×(7y)=5×256x4×7y=1280x4×7yT(2) = 5 \times (4x)^4 \times (7y) = 5 \times 256x^4 \times 7y = 1280x^4 \times 7y.
  5. Final Simplified Form: Multiply the constants to get the final simplified form of the 22nd term. T(2)=1280×7×x4×y=8960x4yT(2) = 1280 \times 7 \times x^4 \times y = 8960x^4y.

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