Question

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

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

Answer 1

Answer: (B), (D)

$\frac{-17}{2}$

$\Rightarrow \underset{n\to \infty }{lim}\frac{(1^a+2^a+...+n^a)}{{(n+1)}^{a-1}\left[\right(na+1)+(na+2)+...+(na+n\left)\right]}=\frac{1}{60}\Rightarrow \underset{n\to \infty }{lim}\frac{\sum _{n=1}^n(r{)}^a}{(n+1{)}^{a-1}\left[\sum _{n=1}^n(na+r)\right]}=\frac{1}{60}\Rightarrow \underset{n\to \infty }{lim}\frac{n^a\sum _{n=1}^n{\left(\frac{r}{n}\right)}^a}{n(n+1{)}^{a-1}\left[\sum _{n=1}^n(a+\frac{r}{n})\right]}=\frac{1}{60}\Rightarrow \underset{n\to \infty }{lim}\frac{\sum _{n=1}^n{\left(\frac{r}{n}\right)}^a}{{(1+\frac{1}{n})}^{a-1}\left[\sum _{n=1}^n(a+\frac{r}{n})\right]}=\frac{1}{60}\Rightarrow \frac{\underset{0}{\overset{1}{\int }}{x}^{a}dx}{\underset{0}{\overset{1}{\int }}(a+x)dx}=\frac{1}{60}\Rightarrow \frac{{\left[x^{a+1}\right]}_0^1}{(a+1)\left[{(ax+\frac{x^2}{2})}_0^1\right]}=\frac{1}{60}\Rightarrow \frac{1}{a+1}\left[\frac{1-0}{a+\frac{1}{2}}\right]=\frac{1}{60}\Rightarrow \frac{2}{(2a+1)(a+1)}=\frac{1}{60}$

Calculating the value of $a.$

$2a^2+3a-119=0\Rightarrow a=\frac{-3\pm \sqrt{9+8\left(119\right)}}{4}\Rightarrow a=\frac{-3\pm \sqrt{961}}{4}\Rightarrow a=7,-\frac{17}{2}$