Question

The value of $n\stackrel{lim}{\to }\infty \frac{1.n+2.(n-1)+3.(n-2)+...+n.1}{1^2+2^2+...+n^2}$ is

• A. $1$
• B. $-1$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Now,

$1.n+2(n-1)+3.(n-2)+....+n.1$

The general term of the above series is

$T_r=r(n-r+1)=r(n+1-r)$

Sum of series$=S={\sum }_{r=1}^n{T}_r={\sum }_{r=1}^{n}r(n+1-r)$

$S={\sum }_{r=1}^{n}nr+{\sum }_{r=1}^{n}r-{\sum }_{r=1}^n{r}^2=n{\sum }_{r=1}^{n}r+{\sum }_{r=1}^{n}r-{\sum }_{r=1}^n{r}^2=n\frac{n(n+1)}{2}+\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}$

Similarly, $1^2+2^2+3^2+.....+n^2=\frac{n(n+1)(2n+1)}{6}$

${lim}_{n\to \infty }\frac{1.n+2(n-1)+3.(n-2)+...+n.1}{1^2+2^2+3^2+........n^2}{lim}_{n\to \infty }\frac{\frac{n^2(n+1)}{2}+\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)(2n+1)}{6}}{lim}_{n\to \infty }\frac{\frac{n}{2}+\frac{1}{2}-\frac{2n+1}{6}}{\frac{2n+1}{6}}{lim}_{n\to \infty }\left(\frac{3n+3-2n-1}{2n+1}\right)={lim}_{n\to \infty }\left(\frac{3+\frac{3}{n}-2-\frac{1}{n}}{2+\frac{1}{n}}\right)=\frac{3-2}{2}=\frac{1}{2}$