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The expansion of (1−3x)ⁿ, if the coefficient of x² is 90, find the value of n.

User Jeff Finn
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Final answer:

To find the value of n for which the coefficient of x² is 90 in the expansion of (1−3x)ⁿ, the binomial theorem is used. The coefficient of x² is found by the binomial coefficient nC2, leading to the equation n² - n - 20 = 0, which has a positive solution of n = 5.

Step-by-step explanation:

To determine the value of n for which the coefficient of is 90 in the expansion of (1−3x)ⁿ, we use the binomial theorem. The binomial theorem gives us a way to expand expressions raised to a power and is written as:

  • (a + b)ⁿ = aⁿ + naⁿ−1b + ⅟n(n-1)aⁿ−62 + ⅟n(n-1)(n−2)aⁿ∓63 + ... + bⁿ

The general term in the expansion of (a + b)ⁿ is given by T(k+1) = nCk × aⁿ−k × b⁾k, where nCk is the binomial coefficient.

To find the coefficient of , we need the term where k=2. In our case, a = 1 and b = -3x, and thus the term becomes:

  • T(3) = nC2 × 1ⁿ−2 × (-3x)² = n(n-1)/2! × 9x²

So, the coefficient of is:

  • 9 × n(n-1)/2 = 90

Simplifying gives us:

  • 9n(n - 1) = 180
  • n(n - 1) = 20
  • n² - n - 20 = 0
  • (n - 5)(n + 4) = 0

Since n is positive, n = 5 is the value of n we're looking for.

User Rahul Shinde
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