In the binomial expansion of (2k x) n , where k is a constant and n is a positive integer, the coefficient of x 2 is equal to the coefficient of x 3 a Prove that n 6k 2 b Given also that k 2 3, expand (2k x) n in ascending powers of x up to and including the term in x 3 , giving each coefficient as...

In the binomial expansion of (2k + x) n , where k is a constant and n

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In the binomial expansion of (2k + x)ⁿ, where k is a constant and n is a positive integer, the coefficient of x² is equal to the coefficient of x³.

a. Prove that n = 6k + 2.

b. Given also that k = 2/3, expand (2k + x)ⁿ in ascending powers of x up to and including the term in x³, giving each coefficient as an exact fraction in its simplest form.

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a To prove that n 6k 2 we need to equate the coefficients of x2 and x3 in the binomial expansion of ...View the full answer

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