A= \(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)
2A= \(2.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\right)\)
2A= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
⇒ 2A- A= \(1-\dfrac{1}{2^{100}}\)
⇒ A= \(1-\dfrac{1}{2^{100}}\)
B= \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
3B= \(3.\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\)
3B= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
⇒ 3B- B= \(1-\dfrac{1}{3^{100}}\)
⇒ B.(3-1)= \(1-\dfrac{1}{3^{100}}\)
⇒ 2B= \(1-\dfrac{1}{3^{99}}\)
⇒ B= \(\left(1-\dfrac{1}{3^{99}}\right):2\)
⇒ B= \(\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)
