B=($\frac{1}{2}$ )+($\frac{1}{2}$ )^2+($\frac{1}{2}$ )^3+($\frac{1}{2}$ )^3+($\frac{1}{2}$ )^4+…+($\frac{1}{2}$ )^98+($\frac{1}{2}$ )^99
B=($\frac{1}{2}$ )+($\frac{1}{2}$ )^2+($\frac{1}{2}$ )^3+($\frac{1}{2}$ )^3+($\frac{1}{2}$ )^4+…+($\frac{1}{2}$ )^98+($\frac{1}{2}$ )^99
Đáp án:
Giải thích các bước giải:
(1/2)B=(1/2)^2+(1/2)^3+……+(1/2)^99+(1/2)^100
B-(1/2)B=(1/2 )+(1/2 )^2+(1/2 )^3+(1/2 )^3+(1/2 )^4+…+(1/2 )^98+(1/2 )^99-[(1/2)^2+(1/2)^3+……+(1/2)^99+(1/2)^100]
1/2B=1/2-(1/2)^100
1/2B=1/2[1-(1/2)^99]
B=1-(1/2)^99
Đáp án:
=> B = $\frac{1}{2}$ + $\frac{1}{2^{2}}$ + $\frac{1}{2^{3}}$ + … + $\frac{1}{2^{99}}$ (1)
=> 2B = 1 + $\frac{1}{2}$ + $\frac{1}{2^{2}}$ + … + $\frac{1}{2^{98}}$ (2)
Lấy (2) – (1) ta đc
B = 1 – $\frac{1}{2^{99}}$ = $\frac{2^{99}-1}{2^{99}}$
Giải thích các bước giải: