Another Coefficients Problem

$(1-x-x^{2}+x^{3})^{6}=(1-x)^{6}(1-x^{2})^{6}=(1-x)^{12}(1+x)^{6}$

Using binomial expansion, one may calculate coefficient of $x^{7}$ as $-\binom{12}{7}\binom{6}{0}+\binom{12}{6}\binom{6}{1}-\binom{12}{5}\binom{6}{2}+\binom{12}{4}\binom{6}{3}-\binom{12}{3}\binom{6}{4}+\binom{12}{2}\binom{6}{5}-\binom{12}{1}\binom{6}{6}$ which equals $\boxed{-144}$

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

$=(1-x)^{6} (1-x^{2})^{6}$

Availing binomial expansion, co-efficient of $x^{7}$

$=(6C1)^{2}-(6C2)(6C3)+(6C3)(6C1)$

$= \boxed {-144}$

• 7=3+3+1 which corresponds to -60
• 7=3+2+2 which corresponds to +60
• 7=3+2+1+1 which corresponds to -180
• 7=3+1+1+1+1 which corresponds to +30
• 7=2+2+2+1 which corresponds to +60
• 7=2+2+1+1+1 which corresponds to -60
• 7=2+1+1+1+1+1 which corresponds to 6

Add them to obtain -144.

$Exp. = (1 - X^2 )^6*(1-X)^6~expanding~ by~ Pascal, \\ taking~ only~terms ~less~than~X^8,~ from ~the~ first~ bracket,~ and\\since~ the~ first~ bracket~ has~ terms ~only~ of~ even~power~ of~ X, \\to~obtain ~X^7, we~ need~ only ~terms~ of~ odd~ power~ of~ X \\from~ the~ second ~bracket..\\$ $Exp.\\=(..-20X^6+15X^4-6X^2 +1)*(..-6X^5 +..-20X^3+..-6X+..)\\ So, ~ multiplying ~terms~ that~ give~ X^7 \\ -20X^6*(-6X) +15X^4*(-20)X^3-6X^2*(-6X^5) \\=(+120-300+36)X^7= \boxed{-144}*X^7$

$(1-x-x^2+x^3 )^6$ let $(1-x)$ be a, $(x^3-x^2)$ be b, by pascal, we get $(a+b)^6 =a^6 +6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$ now change a&b $(1-x)^6+6(1-x)^5 (x^3-x^2 )+15(1-x)^4 (x^3-x^2 )^2+ 20(1-x)^3 (x^3-x^2 )^3+15(1-x)^2 (x^3-x^2 )^4+6(1-x)^1 (x^3-x^2 )^5+(x^3-x^2 )^6$ use pascal again,notice that $(1-x)^6$ doesnt have a 7th power, so exclude it,again after $20(1-x)^3 (x^3-x^2 )^3$ the powers are too big. $6(1-5x+10x^2-10x^3+5x^4-x^5 )(x^3-x^2 )+15(1-4x+6x^2-4x^3+x^4 )(x^6-2x^5+x^4 )+20(1-3x+3x^2-x^3 )(x^9-3x^8+3x^7-x^6 )$ now multiply the powers to get the power( only those whose power add up to 7) $6(6x^7 )+15(-4x^7-12x^7-4x^7 )+20(3x^7+3x^7 )$ $36-300+120$ we get $\boxed {-144}$