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Find the 2 nd term in the expansion of 
(2x+3)^(6) in simplest form.
Answer:

Find the 22 nd term in the expansion of (2x+3)6 (2 x+3)^{6} in simplest form.\newlineAnswer:

Full solution

Q. Find the 22 nd term in the expansion of (2x+3)6 (2 x+3)^{6} in simplest form.\newlineAnswer:
  1. Use Binomial Theorem: To find the 22nd term in the expansion of (2x+3)6(2x+3)^{6}, 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,k1)a(nk+1)b(k1)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: First, we calculate the binomial coefficient for the 22nd term, which is C(6,21)=C(6,1)C(6, 2-1) = C(6, 1). The binomial coefficient C(n,k)C(n, k) is calculated as n!k!(nk)!\frac{n!}{k! \cdot (n-k)!}, where "!" denotes factorial.
  3. Correct Factorial Calculation: Calculating C(6,1)C(6, 1) gives us 6!(1!(61)!)=6(15!)=6(1120)=6120=120\frac{6!}{(1! * (6-1)!)} = \frac{6}{(1 * 5!)} = \frac{6}{(1 * 120)} = \frac{6}{120} = \frac{1}{20}. This is incorrect because we made a mistake in the factorial calculation. The correct calculation should be 6(15!)=6(1120)=6120=120\frac{6}{(1 * 5!)} = \frac{6}{(1 * 120)} = \frac{6}{120} = \frac{1}{20}, which simplifies to 66.

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