The given series is a type of infinite geometric series. The general form of the series is: \( S = 1 - \frac{2}{3} + \frac{2.4}{3.6} - \frac{2.4.6}{3.6.9} + \cdots \)
Step 1: Express this as a geometric series with first term \( 1 \) and common ratio \( \frac{-2}{3} \). The sum of an infinite geometric series is given by: \( S = \frac{a}{1 - r} \) Where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = 1 \) and \( r = -\frac{2}{3} \). \( S = \frac{1}{1 - \left(-\frac{2}{3}\right)} = \frac{1}{1 + \frac{2}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5} \)
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$ 2^m 3^n 5^k, \text{ where } m, n, k \in \mathbb{N}, \text{ then } m + n + k \text{ is equal to:} $
Let $ (1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + ... + a_{20} x^{20} $. If $ (a_1 + a_3 + a_5 + ... + a_{19}) - 11a_2 = 121k $, then k is equal to _______