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

Find the 22nd term in the expansion of (2x−5y)5 (2 x-5 y)^{5} in simplest form.\newlineAnswer:

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Full solution

Q. Find the 22nd term in the expansion of (2x−5y)5 (2 x-5 y)^{5} in simplest form.\newlineAnswer:
  1. Use Binomial Theorem: To find the 22nd term in the expansion of (2x−5y)5(2x-5y)^{5}, 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)⋅an−k+1⋅bk−1T(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(5,2−1)=C(5,1)C(5, 2-1) = C(5, 1). The binomial coefficient C(n,k)C(n, k) is calculated as n!k!⋅(n−k)!\frac{n!}{k! \cdot (n-k)!}, where “!“\text{“!“} denotes factorial.
  3. Calculate Powers of a and b: Calculating C(5,1)C(5, 1) gives us 5!1!∗(5−1)!=51∗4!=51∗24=524\frac{5!}{1! * (5-1)!} = \frac{5}{1 * 4!} = \frac{5}{1 * 24} = \frac{5}{24}. Since 2424 is the factorial of 44, we simplify it to get 524=5\frac{5}{24} = 5.
  4. Multiply Coefficient and Powers: Now we need to calculate the powers of aa and bb for the 22nd term. Since a=2xa = 2x and b=−5yb = -5y, and we are looking for the 22nd term where k=2k=2, we have a5−2+1=a4a^{5-2+1} = a^{4} and b2−1=b1b^{2-1} = b^{1}.
  5. Simplify the Expression: Calculating the powers, we get (2x)4=16x4(2x)^{4} = 16x^{4} and (−5y)1=−5y(-5y)^{1} = -5y.
  6. Simplify the Expression: Calculating the powers, we get (2x)4=16x4(2x)^{4} = 16x^4 and (−5y)1=−5y(-5y)^{1} = -5y.Multiplying the binomial coefficient by the powers of aa and bb, we get the 22nd term: T(2)=C(5,1)⋅(2x)4⋅(−5y)1=5⋅16x4⋅−5yT(2) = C(5, 1) \cdot (2x)^{4} \cdot (-5y)^{1} = 5 \cdot 16x^4 \cdot -5y.
  7. Simplify the Expression: Calculating the powers, we get (2x)4=16x4(2x)^{4} = 16x^4 and (−5y)1=−5y(-5y)^{1} = -5y.Multiplying the binomial coefficient by the powers of aa and bb, we get the 22nd term: T(2)=C(5,1)⋅(2x)4⋅(−5y)1=5⋅16x4⋅−5yT(2) = C(5, 1) \cdot (2x)^{4} \cdot (-5y)^{1} = 5 \cdot 16x^4 \cdot -5y.Simplifying the expression, we get T(2)=5⋅16⋅−5⋅x4⋅y=−400⋅x4⋅yT(2) = 5 \cdot 16 \cdot -5 \cdot x^4 \cdot y = -400 \cdot x^4 \cdot y.

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