Question

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

• A. $\frac{256}{243}$
• B. $\frac{4^4}{6^5}$
• C. $\frac{243}{256}$
• D. $\infty$

Answer 1

The correct option is A $\frac{256}{243}$

$\underset{n\to \infty }{lim}\frac{n(1^2+2^2+3^2+..\cdot .\cdot .\cdot \cdot +n^2{)}^5}{(1^3+2^3+3^3++n^3{)}^4}$
$\underset{n\to \infty }{lim}\frac{n({\sum }_{k=1}^n{k}^2{)}^5}{({\sum }_{k=1}^n{k}^3{)}^4}$
use summation formula
$\underset{n\to \infty }{lim}\frac{n(\frac{n(n+1)(2n+1)}{6}{)}^5}{(\frac{n(n+1)}{2}{)}^8}$
$\underset{n\to \infty }{lim}\frac{n^{16}(\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}{)}^5}{n^{16}(\frac{1+\frac{1}{n}}{2}{)}^8}$
$\underset{n\to \infty }{lim}\frac{(\frac{1}{3}{)}^5}{(\frac{1}{2}{)}^8}=\frac{2^8}{3^5}$