Question

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

• A. $\frac{1}{24}$
• B. $\frac{1}{12}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

Answer: (A)

$\frac{1}{24}$

GIven:${lim}_{x\to \infty }\frac{{\sum }_{k=1}^{n}k+2{\sum }_{k=1}^{n-1}k..........n.1}{n^4}$

To find: The value of the limit

Sol: We can write the above expressions as

${lim}_{x\to \infty }\frac{{\sum }_{k=1}^{n}k(k.{\sum }_{r=1}^{n-k+1}r)}{n^4}$

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

$⟹{\sum }_{k=1}^{n}k.\left\{\frac{n^2+k^2-2nk+3n-3k+2}{2}\right\}={\sum }_{k=1}^{n}k\left\{\frac{n^2+k^2-(2x+3)k+(3x+2)}{2}\right\}$

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

$==\left(\frac{n^2+3n+2}{2}\right)-\frac{n(n+1)}{2}+\frac{1}{2}{\left(\frac{n(n+1)}{2}\right)}^2-\frac{(2n+3)}{2}\frac{n(n+1)(2n+1)}{6}$

$=\left(\frac{n^2+3n+2}{2}\right)-\frac{n(n+1)}{2}+\frac{1}{2}{\left(\frac{n(n+1)}{2}\right)}^2-\frac{(2n+3)}{2}\frac{n(n+1)(2n+1)}{6}$

$=\frac{n(n+1)(n^2+5n+6)}{24}$

$\therefore {lim}_{n\to \infty }\frac{{\sum }_{k=1}^n(k.{\sum }_{r=1}^{n-k+1}r)}{n^4}$

$=\frac{1}{24}{lim}_{n\to \infty }\frac{n(n+1)(n+2)(n+3)}{n^2}$

$=\frac{1}{24}{lim}_{n\to \infty }(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})$

$=\frac{1}{24}.1$

Hence, correct answer is 1/24