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Ta có:
$C_i^3.C_{2012}^i=\frac{i!}{3!(i-3)!}.\frac{2012!}{i!.(2012-i)!}=\frac{2012!}{3!}.\frac{1}{(i-3)!.(2012-i)!}=C_{2012}^3.C_{2009}^{i-3}$.
Do đó:
$\sum_{i=3}^{2012}{C_i^3.C_{2012}^i}=C_{2012}^3\sum_{i=0}^{2009}{C_{2009}^i}=C_{2012}^3.2^{2009}$.
Giả thiết tương đương với:
$C_{2012}^3.2^{2009}=2011.2^{2011}.x$
$\Leftrightarrow x=168505$.
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