Question

For $aϵ1$ (the set of all real numbers), $a\ne -1,{lim}_{n\to \infty }\frac{(1^a+2^a+...+n^a)}{(n+1{)}^{a-1}[(na+1)+(na+2)+...+(na+n)]}=\frac{1}{60}$. Then $a=$

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

Answer 1

Answer: (B), (D)

$\frac{-17}{2}$

$L{t}_{n\to \infty }\frac{{\sum }_{r=1}^n(r{)}^a}{(n+1{)}^{a-1}\left[{\sum }_{r=1}^n\right(na+r\left)\right]}=\frac{1}{60}$

$L{t}_{n\to \infty }\frac{n^a\sum {\left(\frac{r}{n}\right)}^a}{n(n+1{)}^{a-1}\sum (a+\frac{r}{n})}=\frac{1}{60}$

$=L{t}_{n\to \infty }\frac{1}{{(1+\frac{1}{n})}^{a-1}}\frac{1}{n}\frac{{\sum }_{r=1}^n{\left(\frac{r}{n}\right)}^a}{{\sum }_{r=1}^n(a+\frac{r}{n})}=\frac{1}{60}$

$=\frac{{\int }_0^1{x}^a}{{\int }_0^{1}a+x}=\frac{1}{60}$

$\frac{x^{a+1}{|}_0^1}{(a+1)\left[{\begin{array}{c}(ax+\frac{x^2}{2})\end{array}|}_0^1\right]}=\frac{1}{60}$

$\frac{1}{a+1}\left[\frac{1-0}{a+\frac{1}{2}}\right]=\frac{1}{60}$

$\frac{2}{(2a+1)(a+1)}=\frac{1}{60}$

$2a^2+3a+1=120$

$2a^2+3a-119=0$

$a=\frac{-3\pm \sqrt{9+8\left(119\right)}}{4}$

$=\frac{-3\pm \sqrt{961}}{4}$

$=\frac{-3\pm 31}{4}$

$a=7,-\frac{17}{2}$