SATHEE: Indefinite Integration 4 Question 4

Indefinite Integration 4 Question 4

4. For $a \in R$ (the set of all real numbers), $a \neq - 1$ , $lim_{n \to \infty} \frac{(1^{a} + 2^{a} + \dots + n^{a})}{(n + 1)^{a - 1} [(n a + 1) + (n a + 2) + \dots + (n a + n)]} = \frac{1}{60}$ . Then, $a$ is equal to

(a) 5

(b) 7

(d) $\frac{- 17}{2}$

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Answer:

Correct Answer: 4. (b)

Solution:

PLAN Converting Infinite series into definite Integral

$\begin{array}{ll} i.e. & lim_{n \to \infty} \frac{h (n)}{n} \\ \begin{array}{ll} lim_{n \to \infty} \frac{1}{n} \sum_{r = g (n)}^{h (n)} f & \frac{r}{n} = \int f (x) d x \\ lim_{n \to \infty} \frac{g (n)}{n} \end{array} \end{array}$

where, $\frac{r}{n}$ is replaced with $x$ .

$Σ$ is replaced with integral.

$\begin{aligned} lim_{n \to \infty} \frac{1^{a} + 2^{a} + \dots + n^{a}}{(n + 1)^{a - 1} (n a + 1) + (n a + 2) + \dots + (n a + n)} = \frac{1}{60} \\ \Rightarrow lim_{n \to \infty} \frac{\sum_{r = 1}^{n} r^{a}}{(n + 1)^{a - 1} \cdot n^{2} a + \frac{n (n + 1)}{2}} = \frac{1}{60} \end{aligned}$

$\begin{aligned} \Rightarrow lim_{n \to \infty} \frac{2 \sum_{r = 1}^{n} \frac{r}{n}}{1 + {\frac{1}{n}}^{a - 1} \cdot (2 n a + n + 1)} = \frac{1}{60} \\ \Rightarrow lim_{n \to \infty} \frac{1}{n} 2 \sum_{r = 1}^{n} {\frac{r}{n}}^{a} \cdot lim_{n \to \infty} \frac{1}{1 + {\frac{1}{n}}^{a - 1} \cdot 2 a + 1 + \frac{1}{n}} = \frac{1}{60} \\ \Rightarrow 2 \int_{0}^{1} (x^{a}) d x \cdot \frac{1}{1 \cdot (2 a + 1)} = \frac{1}{60} \\ \Rightarrow \frac{2 \cdot {[x^{a + 1}]}_{0}^{1}}{(2 a + 1) \cdot (a + 1)} = \frac{1}{60} \\ ∴ \frac{2}{(2 a + 1) (a + 1)} = \frac{1}{60} \Rightarrow (2 a + 1) (a + 1) = 120 \\ \Rightarrow 2 a^{2} + 3 a + 1 - 120 = 0 \Rightarrow 2 a^{2} + 3 a - 119 = 0 \end{aligned}$

Indefinite Integration 4 Question 4

4. For $a \in R$ (the set of all real numbers), $a \neq - 1$ , $lim_{n \to \infty} \frac{(1^{a} + 2^{a} + \dots + n^{a})}{(n + 1)^{a - 1} [(n a + 1) + (n a + 2) + \dots + (n a + n)]} = \frac{1}{60}$ . Then, $a$ is equal to

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Indefinite Integration 4 Question 4

4. For a∈R (the set of all real numbers), a≠−1, limn→∞(1a+2a+…+na)(n+1)a−1[(na+1)+(na+2)+…+(na+n)]=160. Then, a is equal to

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4. For $a \in R$ (the set of all real numbers), $a \neq - 1$ , $lim_{n \to \infty} \frac{(1^{a} + 2^{a} + \dots + n^{a})}{(n + 1)^{a - 1} [(n a + 1) + (n a + 2) + \dots + (n a + n)]} = \frac{1}{60}$ . Then, $a$ is equal to