Question

$\underset{n\to \infty }{lim}\frac{1^3+2^3+3^3+\dots .+n^3}{3n^4+5n^3+6}=$

• A. $\frac{1}{3}$
• B. $\frac{1}{5}$
• C. $\frac{1}{6}$
• D. $\frac{1}{12}$

Answer 1

The correct option is D $\frac{1}{12}$

$\frac{{\sum }_{k=1}^4{k}^3}{3n^4+5n^3+6}=\frac{(\frac{n(n+1)}{2}{)}^2}{n^4(3+\frac{5}{n}+\frac{6}{n^4})}$
$=\frac{(1+\frac{1}{n}{)}^2}{4(3+\frac{5}{n}+\frac{6}{n^4})}$
$=\frac{1}{12}$
Hence, option 'C' is correct.