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

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

Full solution

Q. Find the 22 nd term in the expansion of (x+4)4 (x+4)^{4} in simplest form.\newlineAnswer:
  1. Identify Binomial Expansion Form: Identify the general form of the binomial expansion.\newlineThe binomial theorem states that (a+b)n(a+b)^n expands to a series of terms of the form C(n,k)a(nk)bkC(n, k) \cdot a^{(n-k)} \cdot b^k, where C(n,k)C(n, k) is the binomial coefficient, nn is the power, and kk is the term number minus one.
  2. Determine Binomial Coefficient: Determine the binomial coefficient for the 22nd term.\newlineThe 22nd term corresponds to k=1k=1 (since we start counting from k=0k=0 for the first term). The binomial coefficient C(n,k)C(n, k) for n=4n=4 and k=1k=1 is C(4,1)C(4, 1) which is 44 choose 11.\newlineC(4,1)=4!1!×(41)!=41=4C(4, 1) = \frac{4!}{1! \times (4-1)!} = \frac{4}{1} = 4
  3. Apply Binomial Theorem: Apply the binomial theorem to find the 22nd term. Using the binomial theorem, the 22nd term is given by C(4,1)×x41×41C(4, 1) \times x^{4-1} \times 4^1. This simplifies to 4×x3×44 \times x^3 \times 4.
  4. Simplify 22nd Term: Simplify the 22nd term.\newlineNow we multiply the constants and simplify the term.\newline4×x3×4=16×x34 \times x^3 \times 4 = 16 \times x^3

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