Question

K =\(\dfrac{1}{2}-\dfrac{1}{2^3}-\dfrac{1}{2^5}-\dfrac{1}{2^7}-.........-\dfrac{1}{2^{99}}\)

Answer 1

4K= \(2-\dfrac{1}{2}-\dfrac{1}{2^3}-\dfrac{1}{2^5}-.....-\dfrac{1}{2^{97}}\)

4K - K = \(2-\dfrac{1}{2^{99}}\)

3K= \(2-\dfrac{1}{2^{99}}\)

K=\(\left(2-\dfrac{1}{2^{99}}\right):3\)

Answer 2

\(4K=2-\dfrac{1}{2}-\dfrac{1}{2^5}-...-\dfrac{1}{2^{97}}\)

\(4K-K=2-\dfrac{1}{2^{99}}\)

\(3K=2-\dfrac{1}{2^{90}}\)

\(K=\left(2-\dfrac{1}{2^{90}}\right):3\)

\(K=\dfrac{2}{3}\)