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

Question

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

Answer: 120

Accepted Answer

$$(1+x+x^2+x^3)^6 = \left(\frac{1-x^4}{1-x}\right)^6 = (1-x^4)^6(1-x)^{-6}$$

$$= (1-6x^4+\cdots)\left[1+{}^6C_1x+{}^7C_2x^2+\cdots\right]$$

Coefficient of $x^4$: $\binom94-6\times\binom61 = 126-6=120$

Solutions · 1