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Binomial theorem for positive integral Index

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Question

The coefficient of x⁴ in the expansion of ${(1 + x + x^{2} + x^{3})}^{6}$ in powers of x is .......

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Solution

The coefficient of x⁴ in the expansion of ${(1 + x + x^{2} + x^{3})}^{6}$
= coefficient of x⁴ in ${(\frac{1 - x^{4}}{1 - x})}^{6}$
$\begin{array}{l} = coefticient of x^{4} in (1 - 6 x^{4}) (1 - x)^{- 6} \\ = coefficient of x^{4} in (1 - 6 x^{4}) [1 +^{6} C_{1} x +^{7} C_{2} x^{2} + \dots] \end{array}$
$=^{9} C_{4} - 6 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} - 6 = 126 - 6 = 120$

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