Find the middle term(s) in the expansion of $(1 + 3 x + 3 x^{2} + x^{3})^{2 n}$ .

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Solution

To find :- Middle term in the expansion of $(1 + 3 x + 3 x^{2} + x^{3})^{2 n}$ .
solution :-
$(1 + 3 x + 3 x^{2} + x^{3})^{2 n} = [(1 + x)^{3}]^{2 n}$ $(∵$ we know $(a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2})$
$= (1 + x)^{6 n}$
Here Power $= 6 n = 2 (3 n) = e v e n$
Here Total no. of terms are odd.
So, There are only one middle term
i.e. ${(\frac{6 n}{2})}^{t h} t e r m = (3 n)^{t h} t e r m$
$T_{3 n}$ is to be find now,
we know formula for general term.
$T_{r + 1} =^{n} C_{r} x^{n - r} . y^{r}$ in expansion of $(x + y)^{n}$ .
$T_{3 n} =^{6 n} C_{(3 n - 1)} x^{(3 n - 1)} ⟺$ final answer

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