Question

Tính \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)

Answer 1

bn chắc đề đúng chứ?chổ (1/2)^99 đó,2 cái liền hả?

Answer 2

đề lấy y hệt từ violympic 

Answer 3

\(\frac{1}{2}\)+\(\frac{1^2}{2^2}\)+\(\frac{1^3}{2^3}\)+...+\(\frac{1^{98}}{2^{98}}\)+\(\frac{1^{99}}{2^{99}}\)

=\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+...+\(\frac{1}{2^{99}}\)

=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{2^2}\)+...+\(\frac{1}{2^{98}}\)-\(\frac{1}{2^{99}}\)  Còn lại tự làm nhá kết quả cuối cùng là 2⁹⁹-1/2⁹⁹

Answer 4

duyệt nốt đi

Answer 5

câu này bữa trc thi gặp nè 

Answer 6

 Đặt S=1/2+(1/2)^2+(1/2)^3+......+(1/2)^98+(1/2)^99+(1/2)^99

        S=(1/2+1/2^2+1/2^3+.....+1/2^98+1/2^99)+1/2^99

Đặt A=1/2+1/2^2+1/2^3+......+1/2^98+1/2^99

=>2A=1+1/2+1/2^2+.......+1/2^97+1/2^98

=>2A-A=(1+1/2+1/2^2+.....+1/2^97+1/2^98)-(1/2+1/2^2+1/2^3+....+1/2^98+1/2^99)

=>A=1-1/2^99

Khi đó S=1-1/2^99+1/2^99=1

Vậy..............=1