Question-77881

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

(1+x)^n =Σ_(r=0) ^n a_r x^r =Σ_(r=0) ^n C_n ^r x^r ⇒a_r =C_n ^r b_r =1+(a_r /a_(r−1) ) =1+(C_n ^r /C_n ^(r−1) ) =1+(((n!)/(r!(n−r)!))/((n!)/((r−1)!(n−r+1)!))) =1+(((r−1)!(n−r+1)!)/(r!(n−r)!)) =1+((n−r+1)/r) =((r+n−r+1)/r) =((n+1)/r) ⇒Π_(r=1) ^n b_r =Π_(r=1) ^n ((n+1)/r) =(n+1)^n Π_(r=1) ^n (1/r) =(n+1)^n (1×(1/2)×....×(1/n)) =(((n+1)^n )/(n!))=(((101)^(100) )/(100!)) ⇒ n=100.and the correct answer is b.

$$\left(\mathrm{1}+{x}\right)^{{n}} =\sum_{{r}=\mathrm{0}} ^{{n}} \:{a}_{{r}} {x}^{{r}} \:=\sum_{{r}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{r}} \:{x}^{{r}} \:\Rightarrow{a}_{{r}} ={C}_{{n}} ^{{r}} \\ $$$${b}_{{r}} =\mathrm{1}+\frac{{a}_{{r}} }{{a}_{{r}−\mathrm{1}} }\:=\mathrm{1}+\frac{{C}_{{n}} ^{{r}} }{{C}_{{n}} ^{{r}−\mathrm{1}} }\:=\mathrm{1}+\frac{\frac{{n}!}{{r}!\left({n}−{r}\right)!}}{\frac{{n}!}{\left({r}−\mathrm{1}\right)!\left({n}−{r}+\mathrm{1}\right)!}} \\ $$$$=\mathrm{1}+\frac{\left({r}−\mathrm{1}\right)!\left({n}−{r}+\mathrm{1}\right)!}{{r}!\left({n}−{r}\right)!}\:=\mathrm{1}+\frac{{n}−{r}+\mathrm{1}}{{r}} \\ $$$$=\frac{{r}+{n}−{r}+\mathrm{1}}{{r}}\:=\frac{{n}+\mathrm{1}}{{r}}\:\Rightarrow\prod_{{r}=\mathrm{1}} ^{{n}} \:{b}_{{r}} \\ $$$$=\prod_{{r}=\mathrm{1}} ^{{n}} \frac{{n}+\mathrm{1}}{{r}}\:=\left({n}+\mathrm{1}\right)^{{n}} \prod_{{r}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{r}} \\ $$$$=\left({n}+\mathrm{1}\right)^{{n}} \left(\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}×….×\frac{\mathrm{1}}{{n}}\right)\:=\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}!}=\frac{\left(\mathrm{101}\right)^{\mathrm{100}} }{\mathrm{100}!}\:\Rightarrow \\ $$$${n}=\mathrm{100}.{and}\:{the}\:{correct}\:{answer}\:{is}\:{b}. \\ $$

Commented by TawaTawa last updated on 12/Jan/20