Question

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

Answer 1

Answer: 3.00

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