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Find the 2nd term in the expansion of (x+8y)^(4) in simplest form. Answer:

Find the 22nd term in the expansion of (x+8y)4 (x+8 y)^{4} in simplest form.\newlineAnswer:

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Full solution

Q. Find the 22nd term in the expansion of (x+8y)4 (x+8 y)^{4} in simplest form.\newlineAnswer:
  1. Use Binomial Theorem: To find the 22nd term in the expansion of (x+8y)4(x+8y)^{4}, we will use the binomial theorem. The general form of the kk-th term in the expansion of (a+b)n(a+b)^{n} is given by T(k)=C(n,k−1)⋅a(n−k+1)⋅b(k−1)T(k) = 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 22nd term, k=2k=2.
  2. Calculate Binomial Coefficient: We calculate the binomial coefficient for the 22nd term, which is C(4,2−1)=C(4,1)C(4, 2-1) = C(4, 1). The binomial coefficient C(n,k)C(n, k) can be calculated as n!k!⋅(n−k)!\frac{n!}{k! \cdot (n-k)!}, where “!“\text{“!“} denotes factorial.
  3. Correct Calculation: Now we calculate C(4,1)=4!(1!∗(4−1)!)=4(1∗3!)=4(1∗6)=46=23C(4, 1) = \frac{4!}{(1! * (4-1)!)} = \frac{4}{(1 * 3!)} = \frac{4}{(1 * 6)} = \frac{4}{6} = \frac{2}{3}. However, this is a mistake because C(4,1)C(4, 1) should be calculated as 4!(1!∗(4−1)!)=4(1∗3!)=4(1∗6)=46=23\frac{4!}{(1! * (4-1)!)} = \frac{4}{(1 * 3!)} = \frac{4}{(1 * 6)} = \frac{4}{6} = \frac{2}{3}, which simplifies to 44, not 23\frac{2}{3}.

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