The coefficient of $$x^4$$ in the expansion of $$(1 + x + x^2 + x^3)^6$$ in powers of $$x$$, is.......

We begin with the observation that the trinomial inside the bracket can be rewritten in a convenient quotient form. Since $$1 + x + x^2 + x^3$$ is a finite geometric series, we have the standard identity

$$1 + x + x^2 + x^3 \;=\; \dfrac{1 - x^{4}}{1 - x}\;.$$

Therefore,

$$(1 + x + x^2 + x^3)^6 \;=\;\left(\dfrac{1 - x^{4}}{1 - x}\right)^{6} \;=\;(1 - x^{4})^{6}\,(1 - x)^{-6}\;.$$

Our task is to find the coefficient of $$x^{4}$$ in this product. We shall expand each factor separately and then combine like terms.

First, we recall the ordinary Binomial Theorem for a positive integer exponent $$n$$:

$$(1 + y)^{n} \;=\;\sum_{k = 0}^{n} \binom{n}{k}\,y^{k}\;.$$

Applying it to $$(1 - x^{4})^{6}$$ with $$y = -x^{4}$$ and $$n = 6$$, we get

$$(1 - x^{4})^{6} \;=\;\sum_{k = 0}^{6} \binom{6}{k}\,(-1)^{k}\,x^{4k}\;.$$

Next, we recall the generalised Binomial Series for a negative integer exponent. For any positive integer $$m$$,

$$(1 - x)^{-m} \;=\;\sum_{r = 0}^{\infty} \binom{m + r - 1}{r}\,x^{r}\;.$$

Here $$m = 6$$, so

$$(1 - x)^{-6} \;=\;\sum_{r = 0}^{\infty} \binom{6 + r - 1}{r}\,x^{r} \;=\;\sum_{r = 0}^{\infty} \binom{5 + r}{5}\,x^{r}\;.$$

We now multiply the two series

$$\left(\sum_{k = 0}^{6} \binom{6}{k}\,(-1)^{k}\,x^{4k}\right) \;\left(\sum_{r = 0}^{\infty} \binom{5 + r}{5}\,x^{r}\right)$$

and collect the coefficient of $$x^{4}$$. A general term arising from the product is

$$\binom{6}{k}\,(-1)^{k}\,x^{4k}\;\cdot\;\binom{5 + r}{5}\,x^{r} \;=\;\binom{6}{k}\,(-1)^{k}\,\binom{5 + r}{5}\,x^{4k + r}\;.$$

For the exponent of $$x$$ to be exactly $$4$$, we must have

$$4k + r = 4\;.$$

Because $$k$$ is an integer between $$0$$ and $$6$$ inclusive, the only permissible values are

• $$k = 0,\; r = 4$$
• $$k = 1,\; r = 0$$

We evaluate each contribution separately.

Case 1 (k = 0, r = 4):

The coefficient from the first series is $$\binom{6}{0}(-1)^{0} = 1$$, and from the second series is $$\binom{5 + 4}{5} = \binom{9}{5} = 126$$. Multiplying, the contribution is

$$1 \times 126 = 126\;.$$

Case 2 (k = 1, r = 0):

The coefficient from the first series is $$\binom{6}{1}(-1)^{1} = -6$$, and from the second series is $$\binom{5 + 0}{5} = \binom{5}{5} = 1$$. Multiplying, the contribution is

$$-6 \times 1 = -6\;.$$

Adding both contributions, the total coefficient of $$x^{4}$$ is

$$126 + (-6) = 120\;.$$

So, the answer is $$120$$.