Question

Value of $L=\underset{n\to \infty }{lim}\frac{1}{n^4}\left[1.\left(\sum _{k=1}^{n}k\right)+2.\left(\sum _{k=1}^{n}k\right)+3.\left(\sum _{k=1}^{n}k\right)+......n.1\right]$

• A. $0.25$
• B. $0.5$
• C. $0.75$
• D. $1$

Answer 1

The correct option is A $0.25$

$L=\underset{h\to \infty }{lim}\frac{1}{n^4}\left[1\sum _{k=1}^n+2\sum _{k=1}^{n}k+.....\right]$

$=\underset{h\to \infty }{lim}\frac{1}{n^4}\left[\frac{n(n+1)}{2}\right(1+2+......n\left)\right]$

$=\underset{h\to \infty }{lim}\frac{1}{n^4}\left[\frac{n(n+1)n(n+1)}{4}\right]$

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

Apply L hospital rule

$=\frac{2(n+1)}{8n}$

$=\frac{2}{8}=\frac{1}{4}$.