Question

If $\underset{n\to \infty }{lim}{\left(\frac{{}^{3n}{C}_n}{{}^{2n}{C}_n}\right)}^{1/n}=\frac{A}{B}$, where $A,B$ are co-prime, then $A+B$ is divisible by

• A. $41$
• B. $43$
• C. $17$
• D. $19$

Answer 1

The correct option is B $43$

Let $L=\underset{n\to \infty }{lim}{\left(\frac{{}^{3n}{C}_n}{{}^{2n}{C}_n}\right)}^{1/n}$

$\Rightarrow \ln L=\underset{n\to \infty }{lim}\frac{1}{n}\ln \left(\frac{{}^{3n}{C}_n}{{}^{2n}{C}_n}\right)\Rightarrow \ln L=\underset{n\to \infty }{lim}\frac{1}{n}\ln \left(\frac{\left(3n\right)!\left(n\right)!}{\left(2n\right)!\left(2n\right)!}\right)\Rightarrow \ln L=\underset{n\to \infty }{lim}\frac{1}{n}\ln \left(\frac{(2n+1)(2n+2)\cdots (2n+n)}{(n+1)(n+2)\cdots (n+n)}\right)\Rightarrow \ln L=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\ln \left(\frac{2+r/n}{1+r/n}\right)\Rightarrow \ln L=\underset{0}{\overset{1}{\int }}\ln \left(\frac{2+x}{1+x}\right)dx\Rightarrow \ln L=\underset{0}{\overset{1}{\int }}(\ln (2+x)-\ln (1+x))dx\Rightarrow \ln L=[(x+2)\ln (x+2)-(x+1)\ln (x+1)-1{]}_0^1\Rightarrow \frac{A}{B}=\frac{27}{16}$

Hence, $A+B=43$