Question

Solve : $\underset{n\to \infty }{lim}\frac{(1+2+3+\dots \dots +n\text{terms}).(1^2+2^2+\dots \dots n\text{terms})}{n(1^3+2^3+...+n\text{terms})}$

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$\frac{2}{3}$

Given: $\underset{n\to \infty }{lim}\frac{(1+2+3+\dots \dots +n\text{terms}).(1^2+2^2+\dots \dots n\text{terms})}{n(1^3+2^3+...+n\text{terms})}$

Useful formula
$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$

$\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$

$\sum _{k=1}^n{k}^3={\left(\frac{n(n+1)}{2}\right)}^2$

$=\underset{n\to \infty }{lim}\frac{\frac{n(n+1)}{2}\frac{n(n+1)(2n+1)}{6}}{n{\left(\frac{n(n+1)}{2}\right)}^2}$

$=\underset{n\to \infty }{lim}\frac{\frac{(2n+1)}{6}}{\left(\frac{n}{2}\right)}$

$=\underset{n\to \infty }{lim}\frac{2(2+\frac{1}{n})}{6}$

$=\frac{4}{6}=\frac{2}{3}$ $[\because n\to \infty \Rightarrow \frac{1}{n}\to 0]$

Hence, option ${}^{\prime }{B}^{\prime }$ is correct.