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$^nC_r+2.^nC_{r-1}+^nC_{r-2}=?$

$\begin{array}{1 1} =^{n+2}C_{r} \\ =^{n+2}C_{r+1} \\ =^{n+1}C_{r+1} \\ =^{n+1}C_{r} \end{array}$

1 Answer

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  • $^nC_r+^nC_{r+1}=^{n+1}C_{r+1}$
$^nC_r+2.^nC_{r-1}+^nC_{r-2}=(^nC_r+^nC_{r-1})+(^nC_{r-1}+^nC_{r-2})$
$\:=^{n+1}C_r\:\:+\:\:\:^{n+1}C_{r-1}$
$=^{n+2}C_{r}$
answered Sep 6, 2013 by rvidyagovindarajan_1
 
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