Question

Find the coefficient of x4 in the expansion of (1 + x + x2 +x3}^6

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

${(1 + x + {x}^{2} + {x}^{3}) }^{6}$

$= { \{(1 + x \: )+ {x}^{2} (1 + {x}) \} }^{6}$

$= { {(1 + x \: )}^{6} (1+ {x}^{2}) }^{6}$

On Expansion

${ {(1 + x \: )}^{6} (1+ {x}^{2}) }^{6}$

$= \bigg \{ \: (1 + \large{ {}^{6} C_1}x + \large{ {}^{6} C_2} {x}^{2} + .. + \large{ {}^{6} C_6} {x}^{6} \: ) \times (1 + \large{ {}^{6} C_1} {x}^{2} + \large{ {}^{6} C_2} {( {x}^{2}) }^{2} + .. + \large{ {}^{6} C_6} {( {x}^{2}) }^{6} \: \bigg \}$

$= \bigg \{ \: (1 + \large{ {}^{6} C_1}x + \large{ {}^{6} C_2} {x}^{2} + .. + \large{ {}^{6} C_6} {x}^{6} \: ) \times (1 + \large{ {}^{6} C_1} {x}^{2} + \large{ {}^{6} C_2} ( {x}^{4}) + .. + \large{ {}^{6} C_6} ( {x}^{12}) \: \bigg \}$

Hence the coefficient of ${x}^{4} \: \:$

$= \large{ {}^{6} C_2} + (\large{ {}^{6} C_2} \times \large{ {}^{6} C_1} )+ \large{ {}^{6} C_4}$

$= 15 + (15 \times 6) + 15$

$= 15 + 90 + 15$

$= 120$

Answer 2

Answer:

$120$

Step-by-step explanation:

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