Question Number 77881 by TawaTawa last updated on 11/Jan/20
Commented by TawaTawa last updated on 11/Jan/20
$$\mathrm{Please}\:\mathrm{help}.\:\:\mathrm{14},\:\mathrm{15},\:\mathrm{16},\:\mathrm{17} \\ $$
Commented by abdomathmax last updated on 12/Jan/20
$$\left(\mathrm{1}+{x}\right)^{{n}} =\sum_{{r}=\mathrm{0}} ^{{n}} \:{a}_{{r}} {x}^{{r}} \:=\sum_{{r}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{r}} \:{x}^{{r}} \:\Rightarrow{a}_{{r}} ={C}_{{n}} ^{{r}} \\ $$$${b}_{{r}} =\mathrm{1}+\frac{{a}_{{r}} }{{a}_{{r}−\mathrm{1}} }\:=\mathrm{1}+\frac{{C}_{{n}} ^{{r}} }{{C}_{{n}} ^{{r}−\mathrm{1}} }\:=\mathrm{1}+\frac{\frac{{n}!}{{r}!\left({n}−{r}\right)!}}{\frac{{n}!}{\left({r}−\mathrm{1}\right)!\left({n}−{r}+\mathrm{1}\right)!}} \\ $$$$=\mathrm{1}+\frac{\left({r}−\mathrm{1}\right)!\left({n}−{r}+\mathrm{1}\right)!}{{r}!\left({n}−{r}\right)!}\:=\mathrm{1}+\frac{{n}−{r}+\mathrm{1}}{{r}} \\ $$$$=\frac{{r}+{n}−{r}+\mathrm{1}}{{r}}\:=\frac{{n}+\mathrm{1}}{{r}}\:\Rightarrow\prod_{{r}=\mathrm{1}} ^{{n}} \:{b}_{{r}} \\ $$$$=\prod_{{r}=\mathrm{1}} ^{{n}} \frac{{n}+\mathrm{1}}{{r}}\:=\left({n}+\mathrm{1}\right)^{{n}} \prod_{{r}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{r}} \\ $$$$=\left({n}+\mathrm{1}\right)^{{n}} \left(\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}×....×\frac{\mathrm{1}}{{n}}\right)\:=\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}!}=\frac{\left(\mathrm{101}\right)^{\mathrm{100}} }{\mathrm{100}!}\:\Rightarrow \\ $$$${n}=\mathrm{100}.{and}\:{the}\:{correct}\:{answer}\:{is}\:{b}. \\ $$
Commented by TawaTawa last updated on 12/Jan/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by TawaTawa last updated on 12/Jan/20
$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by abdomathmax last updated on 16/Jan/20
$${you}\:{are}\:{welcome} \\ $$