JEE Main 2016 — Mathematics Calculus

$$\underset{n\rightarrow \infty }{\mathrm{lim}}{(\frac{(n+1)(n+2)\ldots .3n}{{n}^{2n}})}^{\frac{1}{n}}$$n→∞lim​(n2n(n+1)(n+2)….3n​)n1​

Let $y={(\frac{(n+1)(n+2)\ldots . (n+2n)}{{n}^{2n}})}^{\frac{1}{n}}$y=(n2n(n+1)(n+2)….(n+2n)​)n1​

$$y={(\frac{(n+1)}{n}\cdot \frac{(n+2)}{n}\ldots \ldots \frac{(n+2n)}{n})}^{\frac{1}{n}}$$y=(n(n+1)​⋅n(n+2)​……n(n+2n)​)n1​

$$\mathrm{log}y=\frac{1}{n}[\mathrm{log}(1+\frac{1}{n})+\mathrm{log}(1+\frac{2}{n})+\ldots ..\mathrm{log}(1+\frac{2n}{n})]$$logy=n1​[log(1+n1​)+log(1+n2​)+…..log(1+n2n​)]

$$=\frac{1}{n}\sum _{r=1}^{2n}\mathrm{ln}(1+r/n)$$=n1​r=1∑2n​ln(1+r/n)

Replace $\displaystyle \sum \rightarrow \int$∑→∫ & $\frac{r}{n}\rightarrow x$nr​→x & $\frac{1}{n}\rightarrow dx$n1​→dx

As $n\rightarrow \infty$n→∞

$$\mathrm{log}y= \int _{0}^{2}\mathrm{log}(1+x)dx$$logy=∫02​log(1+x)dx

Integrating by parts, we get,

$$\mathrm{log}y=\int _{0}^{2}1\cdot \mathrm{log}(1+x)dx$$logy=∫02​1⋅log(1+x)dx

$$={(x\cdot \mathrm{log}(1+x))}_{0}^{2}-\int _{0}^{2}\frac{1}{1+x} xdx$$=(x⋅log(1+x))02​−∫02​1+x1​xdx

$$={(x\mathrm{log}(1+x))}_{0}^{2}- \int _{0}^{2}(\frac{1+x}{1+x})dx+ \int _{0}^{2}\frac{1}{1+x}dx$$=(xlog(1+x))02​−∫02​(1+x1+x​)dx+∫02​1+x1​dx

$$={(x\mathrm{log}(1+x))}_{0}^{2}-{(x)}_{0}^{2}+{(\mathrm{log}(1+x))}_{0}^{2}$$=(xlog(1+x))02​−(x)02​+(log(1+x))02​

$$=(2\mathrm{log}3-0)-(2-0)+(\mathrm{log}3-\mathrm{log}1)$$=(2log3−0)−(2−0)+(log3−log1)

$$=3\mathrm{log}3-2$$=3log3−2

Since $\mathrm{log}y =3\mathrm{log}3-2=\mathrm{log}27-\mathrm{log}{e}^{2}$logy=3log3−2=log27−loge2

$$=y=\frac{{e}^{\mathrm{log}27}}{{e}^{2}}=\frac{27}{{e}^{2}}$$=y=e2elog27​=e227​

$=\frac{27}{{e}^{2}}$=e227​.