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Find the 4 th term in the expansion of 
(2x-3)^(6) in simplest form.
Answer:

Find the 44 th term in the expansion of (2x3)6 (2 x-3)^{6} in simplest form.\newlineAnswer:

Full solution

Q. Find the 44 th term in the expansion of (2x3)6 (2 x-3)^{6} in simplest form.\newlineAnswer:
  1. Identify Variables: To find the 44th term in the expansion of (2x3)6(2x-3)^{6}, we will use the binomial theorem. The binomial theorem states that the nnth term in the expansion of (a+b)m(a+b)^{m} is given by the formula C(m,n1)amn+1bn1C(m, n-1) \cdot a^{m-n+1} \cdot b^{n-1}, where C(m,n1)C(m, n-1) is the binomial coefficient, which can be calculated as m!(mn+1)!(n1)!\frac{m!}{(m-n+1)! \cdot (n-1)!}.
  2. Calculate Binomial Coefficient: First, we need to identify aa, bb, and mm in our expression (2x3)6(2x-3)^{6}. Here, a=2xa = 2x, b=3b = -3, and m=6m = 6.
  3. Calculate a(mn+1)a^{(m-n+1)}: Now, we calculate the 4th4^{th} term, which means n=4n = 4. We need to find C(6,41)C(6, 4-1) which is C(6,3)C(6, 3). The binomial coefficient C(6,3)C(6, 3) is calculated as 6!/[3!(63)!]=(654)/(321)=206! / [3! * (6-3)!] = (6*5*4) / (3*2*1) = 20.
  4. Calculate b(n1)b^{(n-1)}: Next, we calculate a(mn+1)a^{(m-n+1)} which is (2x)(64+1)=(2x)3(2x)^{(6-4+1)} = (2x)^{3}. This equals 8x38x^3.
  5. Multiply Coefficients: Then, we calculate b(n1)b^{(n-1)} which is (3)(41)=(3)3=27(-3)^{(4-1)} = (-3)^{3} = -27.
  6. Final Result: Now, we multiply the binomial coefficient by the powers of 'a' and 'b' to get the 44th term: 20×8x3×(27)20 \times 8x^3 \times (-27). This simplifies to 20×216x320 \times -216x^3.
  7. Final Result: Now, we multiply the binomial coefficient by the powers of 'a' and 'b' to get the 44th term: 20×8x3×(27)20 \times 8x^3 \times (-27). This simplifies to 20×216x320 \times -216x^3. Finally, we multiply 2020 by 216-216 to get the coefficient of x3x^3 in the 44th term: 20×216=432020 \times -216 = -4320. So, the 44th term is 4320x3-4320x^3.

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