Definite Integration Ques 123
For $a \in R$ (the set of all real numbers), $a \neq-1$, $\lim _{n \rightarrow \infty} \frac{\left(1^{a}+2^{a}+\ldots+n^{a}\right)}{(n+1)^{a+1}[(n a+1)+(n a+2)+\ldots+(n a+n)]}=\frac{1}{60}$. Then, $a$ is equal to
5
7
(c) $\frac{-15}{2}$
(d) $\frac{-17}{2}$
Show Answer
Answer:
Correct Answer: 123.(b)
Solution:
- PLAN Converting Infinite series into definite Integral
$$ \begin{array}{ll} \text { i.e. } & \lim _{n \rightarrow \infty} \frac{h(n)}{n} \\ \begin{array}{ll} \lim _{n \rightarrow \infty} \frac{1}{n} \sum _{r=g(n)}^{h(n)} f\left(\frac{r}{n}\right)=\int f(x) d x \ & \lim _{n \rightarrow \infty} \frac{g(n)}{n} \end{array} \end{array} $$
where, $\frac{r}{n}$ is replaced with $x$.
$\Sigma$ is replaced with integral.
$$ \begin{aligned} & \lim _{n \rightarrow \infty} \frac{1^{a}+2^{a}+\ldots+n^{a}}{(n+1)^{a-1}{(n a+1)+(n a+2)+\ldots+(n a+n)}}=\frac{1}{60} \\ & \Rightarrow \lim _{n \rightarrow \infty} \frac{\sum _{r=1}^{n} r^{a}}{(n+1)^{a-1} \cdot n^{2} +\frac{n(n+1)}{2}}=\frac{1}{60} \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \lim _{n \rightarrow \infty} \frac{2 \sum _{r=1}^{n} \frac{r}{n}}{1+\left(\frac{1}{n}\right)^{a-1} \cdot(2 n a+n+1)}=\frac{1}{60} \ & \Rightarrow \lim _{n \rightarrow \infty} \frac{1}{n} \cdot 2 \sum _{r=1}^{n} \left(\frac{r}{n}\right)^{a} \cdot \lim _{n \rightarrow \infty} \frac{1}{1+\left(\frac{1}{n}\right)^{a-1} \cdot 2 a+1+\frac{1}{n}}=\frac{1}{60} \ & \Rightarrow 2 \int _0^{1}\left(x^{a}\right) dx \cdot \frac{1}{1 \cdot(2 a+1)}=\frac{1}{60} \\ & \Rightarrow \quad \frac{2 \cdot\left[x^{a+1}\right] _0^{1}}{(2 a+1) \cdot(a+1)}=\frac{1}{(2 a+1)(a+1)} \\ & \therefore \quad \frac{2}{(2 a+1)(a+1)}=\frac{1}{60} \Rightarrow(2 a+1)(a+1)=120 \\ & \Rightarrow \quad 2 a^{2}+3 a+1-120=0 \Rightarrow 2 a^{2}+3 a-119=0 \end{aligned} $$
BETA
