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Q.
The sum of $\displaystyle\sum_{ r =0}^{ n }(-1)^{ r } \frac{{ }^{ n } C _{ r }}{{ }^{ r +2} C _{ r }}$ is equal to
Binomial Theorem
Solution:
$\displaystyle\sum_{r=0}^n(-1)^r \frac{{ }^n C_r}{{ }^{r+2} C_r}=\displaystyle\sum_{r=0}^n(-1)^r \frac{n !}{(n-r) ! r !} \cdot \frac{r ! \cdot 2 !}{(r+2) !}$
$=2 \displaystyle\sum_{ r =0}^{ n }(-1)^{ r } \frac{ n !}{( n - r ) !( r +2) !}=\frac{2}{( n +1)( n +2)} \cdot \sum_{ r =0}^{ n }(-1)^{ r } \frac{( n +2) !}{( n +2)-( r +2) !( r +2) !} $
$=\frac{2}{( n +1)( n +2)} \cdot \displaystyle\sum_{ r =0}^{ n }(-1)^{ r } \cdot{ }^{ n +2} C _{ r +2}$
$=\frac{2}{( n +1)( n +2)} \cdot\left({ }^{ n +2} C _2-{ }^{ n +2} C _3+{ }^{ n +2} C _4-{ }^{ n +2} C _5-{ }^{ n +2} C _{ n +2}\right) \quad \text { add and subtract }{ }^{ n +2} C _0-{ }^{ n +2} C _1 $
$=\frac{2}{( n +1)( n +2)} \cdot\left\{\left({ }^{ n +2} C _0-{ }^{ n +2} C _1+{ }^{ n +2} C _2-{ }^{ n +2} C _3+{ }^{ n +2} C _4-{ }^{ n +2} C _5-{ }^{ n +2} C _{ n +2}\right)-\left({ }^{ n +2} C _0-{ }^{ n +2} C _1\right)\right\}$
$=\frac{2}{( n +1)( n +2)}(0-(1-( n +2)))=\frac{2}{ n +2}$