Question

Evaluate: $\underset{n\to \infty }{lim}\frac{1}{n^4}[1^2+(1^2+2^2)+(1^2+2^2+3^2)+......(1^2+2^2+3^2+....n^2)]$

• A. $\frac{1}{12}$
• B. $\frac{1}{15}$
• C. $\frac{1}{9}$
• D. $\frac{1}{3}$

Answer 1

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

$\underset{n\to \infty }{lim}\frac{1}{n^4}[1^2+(1^2+2^2)+(1^2+2^2+3^2)+......(1^2+2^2+3^2+....n^2)]=\underset{n\to \infty }{lim}\frac{1}{n^4}\sum _{r=1}^n(1^2+2^2+3^2+....r^2)=\underset{n\to \infty }{lim}\frac{1}{n^4}\sum _{r=1}^n\frac{r(r+1)(2r+1)}{6}=\underset{n\to \infty }{lim}\frac{1}{n^4}\sum _{r=1}^n\frac{2r^3+3r^2+r}{6}=\underset{n\to \infty }{lim}\frac{1}{n^4}[\frac{1}{12}{n^2(n+1)}^2+\frac{1}{12}n(n+1)(2n+1)+\frac{1}{12}n(n+1)]=\underset{n\to \infty }{lim}[\frac{1}{12}{\left(\frac{n+1}{n}\right)}^2+\frac{1}{12}.\frac{1}{n}(1+\frac{1}{n})(2+\frac{1}{n})+\frac{1}{12}.\frac{1}{n^2}(1+\frac{1}{n})]=\frac{1}{12}$