Question

Tính:

P=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

Q=\(4.5^{100}\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\right)+1\)

K=\(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{49.50.51.52}\)

Answer 1

2P=\(\dfrac{2}{2}+\dfrac{2}{2^2}+...+\dfrac{2}{2^{100}}\)

2P=\(1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)

2P-P=\(\dfrac{1}{2}-\dfrac{1}{2^{100}}\)

P=\(\dfrac{1}{2}-\dfrac{1}{2^{100}}\)

Answer 2

\(P=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

\(2P=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)\(\)

\(2P-P=1-\dfrac{1}{2^{100}}\)

\(P=\dfrac{2^{100}}{2^{100}}-\dfrac{1}{2^{100}}\)

\(P=\dfrac{2^{100}-1}{2^{100}}\)