Question

For $a\in R$ (the set of all real numbers),$a\ne -1$,$\underset{\to \infty }{\lim }\frac{(1^a+2^a+3^a+.....+n^a)}{{(n+1)}^{a-1}\left[\right(na+1)+(na+2)+....+(na+n\left)\right]}=\frac{1}{60}$. Then $a=$

• A. $5$
• B. $7$
• C. $\frac{-15}{2}$
• D. $\frac{-17}{2}$

Answer 1

The correct option is B

$7$

Explanation for the correct option:

Step.1 Finding sum individual

Given series, $\underset{\to \infty }{\lim }\frac{(1^a+2^a+3^a+.....+n^a)}{{(n+1)}^{a-1}\left[\right(na+1)+(na+2)+....+(na+n\left)\right]}=\frac{1}{60}......\left(i\right)$

Sum of numerator

$\begin{array}{rcl}& & (1^a+2^a+3^a+.....+n^a)\\ & =& {\int }_0^n{x}^{}dx\\ & =& \frac{n^{a+1}}{a+1}\end{array}$

Sum of denominator

$$\begin{gathered} S_n=\left[\right(na+1)+(na+2)+....+(na+n\left)\right] \\ {S}_n=\frac{n}{2}(2na+n+1) \\ {S}_n=\frac{(2a+1)n^2+n}{2} \end{gathered}$$

Step2. Finding limit of the series

Now, put all the value in equation $\left(i\right)$

$\begin{array}{rcl}\underset{n\to \infty }{\lim }\frac{(1^a+2^a+3^a+.....+n^a)}{{(n+1)}^{a-1}\left[\right(na+1)+(na+2)+....+(na+n\left)\right]}& =& \frac{1}{60}\\ \underset{n\to \infty }{\lim }\frac{{\displaystyle \frac{n^{a+1}}{a+1}}}{{\displaystyle \frac{{(n+1)}^{a-1}\left[\right(2a+1)n^2+n]}{2}}}& =& \frac{1}{60}\\ \frac{2}{a+1}\underset{n\to \infty }{\lim }\frac{{\left(\frac{n}{n+1}\right)}^{a+1}}{(2a+1){\left(\frac{n}{n+1}\right)}^2+{\left(\frac{n}{n+1}\right)}^2}& =& \frac{1}{60}\\ & & \end{array}$

Now,

$\begin{array}{rcl}& & \underset{n\to \infty }{\lim }\frac{n}{1+n}\\ & =& \underset{n\to \infty }{\lim }\frac{n}{n(1+{\displaystyle \frac{1}{n}})}\\ & =& \underset{n\to \infty }{\lim }\frac{1}{1+{\displaystyle \frac{1}{n}}}\\ & =& 1\end{array}$

Using above

$\begin{array}{rcl}\frac{2}{a+1}\underset{n\to \infty }{\lim }\frac{{\left(\frac{n}{n+1}\right)}^{a+1}}{(2a+1){\left(\frac{n}{n+1}\right)}^2+{\left(\frac{n}{n+1}\right)}^2}& =& \frac{1}{60}\\ \frac{2}{a+1}\times \frac{1^{a+1}}{(2a+1)1^2+1^2}& =& \frac{1}{60}\\ (a+1)(2a+1)& =& 120\\ a& =& 7or\frac{-17}{2}\\ & & \end{array}$

Since, $a<0,diverges,a=7$

Hence, correct option is $\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{.}$