Question

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

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{2}$

The numerator is the sum of the cube of first n natural numbers which is $\frac{n^2{(n+1)}^2}{4}$ and the denominator contains the sum of first n natural numbers multiplied by $n^2$.

So the limit becomes $\underset{n\to \infty }{lim}\frac{\frac{n^2{(n+1)}^2}{4}}{\frac{n^{2}n(n+1)}{2}}$

$=\underset{n\to \infty }{lim}\frac{n^2{(n+1)}^2}{4}\times \frac{2}{n^{2}n(n+1)}=\underset{n\to \infty }{lim}\frac{n+1}{2n}$

$=\underset{n\to \infty }{lim}\frac{1}{2}+\frac{1}{2n}=\frac{1}{2}$

Ans- Option $B$