Question

Find the middle term in the expansion of (1+3x+3x^2+x^3)^6

Answer 1

Answer:

The required answer is $T_{10}= { }^{\mathrm{18}} \mathrm{C}_{\mathrm{9}}x^{9}$

Step-by-step explanation:

Middle terms: The middle terms depend on the value of $n$ and are an expansion of $(x + y)^{n}$.

Given: $(1+3x+3x^2+x^3)^6$

To find: the middle term in the expansion of (1+3x+3x^2+x^3)^6

Here, $(1+3x+3x^2+x^3)^6$$=[(1+x)^{3}]^{6}=(1+x)^{18}$

Since $n=18$ is even, so that here $[\frac{1}{2}(18)+1]th$ i.e., 10th term will be middle term.

Where T represents the term

Also, middle term=$T_{10}= { }^{\mathrm{18}} \mathrm{C}_{\mathrm{9}}x^{9}$

Hence,  the middle term in the expansion is $T_{10}= { }^{\mathrm{18}} \mathrm{C}_{\mathrm{9}}x^{9}$

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