Question

Thực hiện phép tính sau:

1, B = \(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+..........+\(\dfrac{1}{2^{2017}}\)

2,\(\dfrac{12}{1.2}\).\(\dfrac{2^2}{2.3}\).\(\dfrac{3^2}{3.4}\).\(\dfrac{4^2}{4.5}\).\(\dfrac{5^2}{5.6}\)

Giúp mình nhé dấu (.) là nhân nha

Answer 1

1,

B=\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.........+\(\dfrac{1}{2^{2017}}\)

2B=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^{2016}}\)

2B-B=(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^{2016}}\))-(\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.......+\(\dfrac{1}{2^{2017}}\))

B=1-\(\dfrac{1}{2^{2017}}\)

Vậy B=1-\(\dfrac{1}{2^{2017}}\)