Find the middle term in the expansion of (1 + 3x + 3x2 + x3)2n

Asked by Aaryan | 1 year ago |  172

##### Solution :-

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 $$\dfrac{ (6n)!}{(3n!)^2} x^{3n}$$

Answered by Sakshi | 1 year ago

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