First slide

Introduction to ITF

Question

$I f n \in N, \sum_{k = 1}^{n} \sin^{- 1} (x_{_{K}}) = \frac{n π}{2}$ then $\sum_{k = 1}^{n} x_{_{K}} =$

Moderate

A

$n$
B

$k$
C

$\frac{k (k + 1)}{2}$
D

$\frac{n (n + 1)}{2}$

Solution

$\begin{array}{l} \sum_{k = 1}^{n} \sin^{- 1} (x_{k}) = \frac{n π}{2} \Rightarrow \sin^{- 1} (x_{k}) = \frac{π}{2} f o r k = 1, 2, . . . n \\ \Rightarrow x_{k} = \sin \frac{π}{2} = 1 f o r k = 1, 2, . . . n \\ ∴ \sum_{k = 1}^{n} x_{k} = 1 + 1 + 1 + ...... + 1 (n t i m e s) = n \end{array}$

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