39now writing the formula of n_{C}_{r} and n+1_{C}_{r+1} and appropriatel;y simplifying we get
k^{2}-2k-7=\frac{r+1}{n+1}
now , we have \frac{r+1}{n+1}\epsilon [\frac{1}{n+1},1]
so, k^{2}-2k-7\epsilon [\frac{1}{n+1},1]
i think u can proceed now
3 Answers
3939sorry.....
the above i had done a mistake...
\frac{r+1}{n+1}\epsilon [\frac{1}{n+1},1]
\frac{r+1}{n+1} can never be 0.