# if

## $\dpi{100} \small \lim_{n \to \infty }\frac{1^{a}+2^{a}+......+n^{a}}{\left ( n+1 \right )^{a-1}\left [ \left ( na+1 \right )+\left ( na+2 \right ) +.......+\left ( na+n \right )\right ]}=\frac{1}{60}$

for some positive real number a, then a is equal to :
Option: 1 7

Option: 2 8

Option: 3 $\frac{15}{2}$

Option: 4 $\frac{17}{2}$

$\large \lim _{n \rightarrow \infty} \frac{1^{a}+2^{a}+3^{a}+\cdots+n^{a}}{(n+1)^{a-1}[(n a+1)+(n a+2)+\cdots+(n a+n)]}=\frac{1}{60}$

$\large \Rightarrow \lim _{n \rightarrow \infty} \frac{1^{a}+2^{a}+\cdots+n^{a}}{(n+1)^{a-1}\left[n^{2} a+\frac{n(n+1)}{2}\right]}=\frac{1}{60}$

$\large \Rightarrow \lim _{n \rightarrow \infty} \frac{2\left[ \sum_{k=1}^{n} k^{a}\right]}{(n+1)^{a-1}\left[2 n^{2} a+n^{2}+n\right]}=\frac{1}{60}$

$\large \lim _{n \rightarrow \infty} \frac{2 \sum_{k=1}^{n}\left(\frac{k}{n}\right)^{a}}{\frac{(n+1)^{a-1}}{n^{a-1}}\left[\frac{2 n^{2} a+n^{2}+n}{n}\right]}=\frac{1}{60}$

$\large \lim _{n \rightarrow \infty} \frac{2 \sum_{k=1}^{n}\left(\frac{k}{n}\right)^{a}}{\left(1+\frac{1}{n}\right)^{a-1}[2 n a+n+1]}=\frac{1}{60}$

$\large \lim _{n \rightarrow \infty} \frac{\frac{2}{n} \sum_{k=1}^{n}\left(\frac{k}{n}\right)^{a}}{\left(1+\frac{1}{n}\right)^{a-1}\left[2 a+1+\frac{1}{n}\right]}=\frac{1}{60}$

$\large {\color{Blue} \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{1}^{n}\left(\frac{k}{n}\right)^{a}=\int_{0}^{1} x^{a} d x \quad[\text { Derivative as limit of a sum }]}$

$\large \therefore \frac{2 \int_{0}^{1} x^{a} d x}{2 a+1}=\frac{1}{60}$

$\large \\\Rightarrow \frac{\left.2\left(x^{a+1}\right)\right]_{0}^{1}}{(a+1)(2 a+1)}=\frac{1}{60} \\ \\\Rightarrow \frac{2}{(a+1)(2 a+1)}=\frac{1}{60} \\ \\\Rightarrow(a+1)(2 a+1)=120 \\ \\\Rightarrow 2 a^{2}+3 a+1=120 \\ \\\Rightarrow 2 a^{2}-3 a=119=0 \\ \\\Rightarrow a=\frac{-3 \pm \sqrt{9+952}}{4}$

$\large =\frac{-3 \pm \sqrt{961}}{4}=\frac{-3 \pm 31}{4} \quad(\text { since } a>0)$

$\large \\\text { Hence, the value of } a \text { is }\\ \\\frac{-3+31}{4}=\frac{28}{4}=7 \Rightarrow a=7$

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