The Taylor series of \( f(x) = \sin^{-1} x \) around \( x = 0 \) is known, and we can square this series to find the Taylor series of \( (f(x))^2 \). The coefficient \( a_6 \) corresponds to the term involving \( x^6 \) in this expanded series. Through a detailed expansion, the value of \( a_6 \) can be calculated, and multiplying this by 9 gives the result: \[ 9a_6 \approx 1.50. \] Thus, the correct answer is \( 1.50 \)
Let $ a_1, a_2, a_3, \ldots $ be in an A.P. such that $$ \sum_{k=1}^{12} 2a_{2k - 1} = \frac{72}{5}, \quad \text{and} \quad \sum_{k=1}^{n} a_k = 0, $$ then $ n $ is:
The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to