We have,
(1 + 3x + 3x2 + x3)2n = (1 + x)6n where, n is an even number.
So the middle term is (\( \dfrac{n}{2}\) + 1)
= (\( \dfrac{6n}{2}\) + 1) = (3n + 1)th term.
Now,
T2n = T3n+1
= 6nC3n x3n
=\( \dfrac{ (6n)!}{(3n!)^2} x^{3n}\)
Hence, the middle term is \( \dfrac{ (6n)!}{(3n!)^2} x^{3n}\)
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