Discussion Forum

1)

 For a ε R ( the set of all real numbers ), a ≠ -1, 

$\lim_{n \rightarrow \infty}\frac{(1^{a}+2^{a}+....+n^{a})}{(n+1)^{a-1}[(na+1)+(na+2)+....+(na+n)]}$

=  $\frac{1}{60}$  . Then ,a is equal to

 

Answer:

Option B,D

Explanation:

Concept involved

 Converting infinite series into definite intergral

 ie     $\lim_{n\rightarrow \infty}\frac{h(n)}{n}$

 $\lim_{n \rightarrow \infty}\frac{1}{n} \sum_{r=g(n)}^{h(n)}f(\frac{r}{n})=\int_{}^{} f(x) dx$

$\lim_{n \rightarrow \infty}\frac{g(n)}{n}$

 where,r/n  is replaced with x

$\sum$  is replaced with integral

 Here  

$\lim_{n \rightarrow \infty}\frac{(1^{a}+2^{a}+....+n^{a})}{(n+1)^{a-1}[(na+1)+(na+2)+....+(na+n)]}$

=  $\frac{1}{60}$

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

$\Rightarrow \lim_{n \rightarrow \infty}\frac{2\sum_{r=1}^{n}(\frac{r}{n})^{a}}{(1+\frac{1}{n})^{a-1}.(2na+n+1)}$

$\Rightarrow \lim_{n \rightarrow \infty}\frac{1}{n}\left(2\sum_{r=1}^{n}(\frac{r}{n})^{a}\right)$

                $\times  \lim_{n \rightarrow \infty}\frac{1}{(1+\frac{1}{n})^{a-1}(2a+1+\frac{1}{n})}$

 $\Rightarrow  2\int_{0}^{1} (x^{a})dx.\frac{1}{1.(2a+1)}$

  $\Rightarrow \frac{2.(x^{a+1})_0^1}{(2a+1).)(a+1)}=\frac{2}{(2a+1)(a+1)}$

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

  $\Rightarrow(2a+1)(a+1)=120$

 $\Rightarrow2a^{2}+3a+1-120=0$

$\Rightarrow2a^{2}+3a-119=0$

 $\Rightarrow$     $(2a+17) (a-7)=0$

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