Question

If $\underset{n\to \infty }{lim}\frac{(1^2+2^2+.......+n^2)(1^4+2^4+.......+n^4)}{(1^7+2^7+....+n^7)}=\frac{K+1}{15};$ then K is equal to

• A. 7
• B. 8
• C. 9
• D. 6

Answer 1

The correct option is A 7

$\underset{n\to \infty }{lim}\frac{(1^2+2^2+...n^2)(1^4+2^4+...n^4)}{(1^7+2^7+...+n^7)}=\underset{n\to \infty }{lim}\frac{{\sum }_{r=1}^n{r}^2{\sum }_{r=1}^n{r}^4}{{\sum }_{r=1}^n{r}^7}$
$=\underset{n\to \infty }{lim}\frac{\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^2\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^2}{\frac{1}{n}{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^6}=\frac{{\int }_0^1{x}^{2}dx{\int }_0^1{x}^{4}dx}{{\int }_0^1{x}^{7}dx}$
$=\frac{{\left[\frac{x^3}{3}\right]}_0^1{\left[\frac{x^5}{5}\right]}_0^1}{{\left[\frac{x^8}{8}\right]}_0^1}=\frac{\left[\frac{1}{3}\right].\left[\frac{1}{5}\right]}{\left[\frac{1}{8}\right]}=\frac{8}{15}=\frac{7+1}{15}$

$\therefore K=7$