AIEEE 2011 (Offline)Question 1
The coefficient of $x^{7}$ in the expansion of $\left(1-x-x^{2}+x^{3}\right)^{6}$ is
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. $1-x-x^2+x^3$ factorises, so the sixth power is a product of two binomials. Formulas used. (1) $1-x-x^2+x^3=(1-x)-x^2(1-x)=(1-x)(1-x^2)$Step 1 - Rewrite. By (1), the expression is $(1-x)^6(1-x^2)^6$. Step 2 - Condition. Take $x^a$ from $(1-x)^6$ and $x^{2b}$ from $(1-x^2)^6$ with $a+2b=7$, $a\le6$: $(a,b)=(5,1),(3,2),(1,3)$. Step 3 - Each case. The term is ${}^{6}C_a(-1)^a\cdot{}^{6}C_b(-1)^b$: $$(5,1):\ (-6)(-6)=36,\qquad(3,2):\ (-20)(15)=-300,\qquad(1,3):\ (-6)(-20)=120.$$ Step 4 - Add. $36-300+120=-144$. Answer: (B)