Question

If ${(1+x)}^n=\sum _{r=0}^n{a}_r{x}^{r}and{b}_r=1+\frac{a_r}{a_{r-1}}and\prod _{r=1}^n{b}_r=\frac{{\left(101\right)}^{100}}{100!}$, then n equals to

• A. 99
• B. 100
• C. 101
• D. none of these

Answer 1

The correct option is B 100

Here,$a_r={{}^{n}C}_r$

$b_r=1+\frac{a_r}{a_{r-1}}$

$=1+\frac{{{}^{n}C}_r}{{{}^{n}C}_{r-1}}$

$=1+\frac{n-r+1}{r}=\frac{r+n-r+1}{r}=\frac{n+1}{r}$

$\Rightarrow {\prod }_{r=1}^n{b}_r$

$={\prod }_{r=1}^n\frac{n+1}{r}$

$=\left(\frac{(n+1)}{1}\right)\left(\frac{(n+1)}{2}\right)\left(\frac{(n+1)}{3}\right)....\left(\frac{(n+1)}{n}\right)=\frac{{(n+1)}^n}{n!}$

$=\frac{{101}^{100}}{100!}$(given)

Hence,$n=100.$