Question

Let $t_{100}=\displaystyle\sum_{r=1}^{100} \frac{1}{\left({ }^{100} C_t\right)^5}$ and $S_{100}=\displaystyle\sum_{r=0}^{100} \frac{r}{\left({ }^{100} C_t\right)^5}$, then the value of $\sec \left(\cos ^{-1}\left(\frac{S_{100}-50}{150 t_{100}}\right)\right)$ equals

Answer 1

${l}x^2 y-2 x y^2-4 x y=0 \Rightarrow x y(x-2 y-4)=0 $
$ r=\frac{\Delta}{s}=\frac{\frac{1}{2} \times 2 \times 4}{3+\sqrt{5}}=\frac{4}{3+\sqrt{5}} $
$=\frac{2}{r}=\frac{2(3+\sqrt{5})}{4}=\frac{6+2 \sqrt{5}}{4}=\frac{(\sqrt{5}+1)^2}{4}$We have,
$S_{100}=\frac{0}{\left({ }^{100} C _0\right)^5}+\frac{1}{\left({ }^{100} C _1\right)^5}+\frac{2}{\left({ }^{100} C _2\right)^5}+\ldots+ $
$\frac{100}{\left({ }^{100} C _{100}\right)^5} \ldots . .(1)$
Also,
$S _{100}=\frac{100}{\left({ }^{100} C _0\right)^5}+\frac{(100-1)}{\left({ }^{100} C _1\right)^5}+\frac{(100-2)}{\left({ }^{100} C _2\right)^5}+ $
$\ldots+\frac{0}{\left({ }^{100} C _{100}\right)^5} \ldots . .(2)\left(\sum_{r=a}^b f(r)=\sum_{r=a}^b f(a+b-r)\right)$
$\therefore$ On adding (1) and (2), we get
$2 S _{100}=100 t _{100}+100 \Rightarrow S _{100}-50=50 t _{100} $
$\frac{ S _{100}-50}{150 t _{100}}=\frac{1}{3}$
Hence, $\sec \left(\cos ^{-1}\left(\frac{S_{100}-50}{150 t_{100}}\right)\right)=3$