In the expansion of (1 – 2x)^n , the sum of the coefficients of x and x^ 2 is 30. Given that n is positive, find (i) the value of n (ii) the coefficient of x
Question
In the expansion of (1 – 2x)^n , the sum of the coefficients of x and x^ 2 is 30. Given that n is positive, find (i) the value of n (ii) the coefficient of x
Solution
(i) The general term in the expansion of (1 – 2x)^n is given by:
T(r+1) = nCr * (1)^(n-r) * (-2x)^r
where nCr = n! / [r!(n-r)!]
The coefficient of x is given when r=1, and the coefficient of x^2 is given when r=2.
So, we have:
Coefficient of x = nC1 * (1)^(n-1) * (-2)^1 = -2n
Coefficient of x^2 = nC2 * (1)^(n-2) * (-2)^2 = 2n(n-1)
Given that the sum of these coefficients is 30, we have:
-2n + 2n(n-1) = 30
Solving this equation gives n = 5.
(ii) The coefficient of x is -2n, so substituting n = 5 gives a coefficient of -10.
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