Question

$y^r=\frac{n!{\cdot }^{n+r-1}{C}_{r-1}}{r^n}$ where $n=2r$. If
$\underset{n\to \infty }{lim}y$ is equal to $\left(\frac{a}{e^b}\right)$ then value of $a+b$ is

Answer 1

$y^r=\frac{n!\cdot (n+r-1)!}{(r-1)!\cdot n!{\dot{r}}^n}$

$y^r=\frac{(n+r-1)(n+r-2)\dots r}{r^n}$

$y^r=\frac{(r+1)(r+2)\dots (n+r-1)}{r^{n-1}}$

$y^r=(1+\frac{1}{r})(1+\frac{2}{r})\dots (1+\frac{n-1}{r})$

$y^r={\prod }_{\alpha =1}^{n-1}(1+\frac{\alpha }{r})$

$\underset{n\to \infty }{lim}\ln y=\underset{n\to \infty }{lim}\frac{1}{r}{\sum }_{\alpha =1}^{n-1}\ln (1+\frac{\alpha }{r})$

$\Rightarrow \ln y={\int }_0^2\ln (1+x)dx$


$=\left[\right(1+x\left)\ln \right(1+x)-(1+x){]}_0^2$

$\Rightarrow \ln y=(3\ln 3-3)-(-1)=3\ln 3-2$

$\Rightarrow \ln y=\ln \left(\frac{27}{e^2}\right)$

$y=\frac{27}{e^2}$
so $a+b=29$