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Find the 2nd term in the expansion of 
(x+4)^(3) in simplest form.
Answer:

Find the 22nd term in the expansion of (x+4)3 (x+4)^{3} in simplest form.\newlineAnswer:

Full solution

Q. Find the 22nd term in the expansion of (x+4)3 (x+4)^{3} in simplest form.\newlineAnswer:
  1. Use Binomial Theorem: To find the 22nd term in the expansion of (x+4)3(x+4)^{3}, we can use the binomial theorem, which states that the nnth term in the expansion of (a+b)n(a+b)^{n} is given by nC(k1)ank+1bk1nC(k-1) \cdot a^{n-k+1} \cdot b^{k-1}, where nC(k1)nC(k-1) is the binomial coefficient. For the 22nd term, k=2k=2.
  2. Calculate Binomial Coefficient: First, we calculate the binomial coefficient for the 22nd term, which is 3C13C1 (since k1=21=1k-1 = 2-1 = 1). 3C13C1 is the number of ways to choose 11 item from 33, which is 33.
  3. Raise x to Power: Next, we raise the first term, xx, to the power of (32+1)(3-2+1), which is x(31)=x2x^{(3-1)} = x^2.
  4. Raise 44 to Power: Then, we raise the second term, 44, to the power of (21)(2-1), which is 41=44^1 = 4.
  5. Multiply Coefficients and Powers: Now, we multiply the binomial coefficient by the powers of xx and 44 to get the 22nd term: 3×x2×43 \times x^2 \times 4.
  6. Simplify Expression: Simplify the expression by multiplying the constants: 3×4=123 \times 4 = 12.
  7. Final 22nd Term: Finally, we have the 22nd term in the expansion: 12x212x^2.

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