In the binomial expansion [tex](2k+x)^{n}[/tex], where k is a constant and n is positive integer, the coefficient of x² is equal to the coefficient of x³(adsbygoogle = window.adsbygoogle || []).push({});

a) Prove that n = 6k + 2

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

my shot at (a)

expand to get x² and x³:

[tex]\stackrel{n}{2}(2k)^{n-2}[/tex] + [tex]\stackrel{n}{3}(2k)^{n-3}[/tex]

subst n = 6k + 2 but i get into more expansion, which i dont think is really going anywhere

can someone help me out/guide me through (a) please?

Thankyou

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# Binomial Theorem expansion with algebra

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