Question

$\underset{n\to \infty }{lim}\frac{(1^4+2^4+3^4+....+(2n{)}^4\left)\right(1^{3/2}+2^{3/2}+....+n^{3/2})}{(1^5+2^5+....+n^5)(1^{1/2}+2^{1/2}+...+n^{1/2})}$

• A. $\frac{48}{5}$
• B. $\frac{18}{25}$
• C. $\frac{576}{25}$
• D. $\frac{128}{5}$

Answer 1

Answer: (C)

$\frac{576}{25}$

${lim}_{x\to \infty }\frac{\left[1^4+2^4+....+{\left(2n\right)}^4\right]\left[1^{\frac{3}{2}}+2^{\frac{3}{2}}+......+n^{\frac{3}{2}}\right]}{\left[1^5+2^5+....+n^5\right]\left[1^{\frac{1}{2}}+2^{\frac{1}{2}}+.....+n^{\frac{1}{2}}\right]}=\frac{n^2\times n^4\times n^{\frac{3}{2}}\left[\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^4\right]\times \left[\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^{\frac{3}{2}}\right]}{n^2\times n^5\times n^{\frac{1}{2}}\left[\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^5\right]\times \left[\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^{\frac{1}{2}}\right]}=\frac{\left({\int }_0^2{x}^{4}dx\right)\left({\int }_0^1{x}^{\frac{3}{2}}dx\right)}{\left({\int }_0^1{x}^{5}dx\right)\left({\int }_0^1{x}^{\frac{1}{2}}dx\right)}=\frac{\frac{2^6}{5^2}}{\frac{1}{3^2}}=\frac{64\times 9}{25}=\frac{576}{25}$

$Hence,theoptionCisthecorrectanswer.$