Binomial Theorem
Coefficient Extraction in Product of Series
nta_pyq_2024_jan
Grade 11

Question:

The coefficient of $x^{2012}$ in the expansion of $(1-x)^{2008}(1+x+x^2)^{2007}$ is equal to

Step-by-Step Solution

Key Concept: Factor: $(1-x)(1+x+x^2)=(1-x^3)$, so $(1-x)^{2008}(1+x+x^2)^{2007}=(1-x)(1-x^3)^{2007}$. For $x^{2012}$ to appear, need $3r=2012$ or $3r+1=2012$ — neither has integer solution.
$(1-x)^{2008}(1+x+x^2)^{2007} = (1-x)(1-x^3)^{2007}$. General term: $(-1)^r\binom{2007}{r}x^{3r}$ and $-(-1)^r\binom{2007}{r}x^{3r+1}$. For $x^{2012}$: need $3r=2012$ (no integer solution) or $3r+1=2012\Rightarrow 3r=2011$ (no integer solution). Hence coefficient of $x^{2012}=0$.
Correct Answer: 0

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