Cho Sn=(1-1/2)+(1-1/4)+(1-1/8)+...+(1-1/2^n). Tính S10

Ta có Sn=n−12+122+123+...+12n

Đặt M=12+122+123+...+12n⇒2M=1+12+122+...+12n−1⇒2M−M=M=1+12+122+...+12n−1−12+122+123...+12n=1−12n⇒Sn=n−1+12n⇒S10=10−1+1210=9+1210