1,Tính. A=(1-2+2^2-2^3+…+2^100-2^101):(1-2^102). Mn g 07/11/2021 Bởi Samantha 1,Tính. A=(1-2+2^2-2^3+…+2^100-2^101):(1-2^102). Mn giúp tui Ik tui đang cần gấp để chiều đi học. Cảm ơn mn nhiều.
Tham khảo `A=1-2+2^2-2^3+…+2^{100}-2^{101}` `⇒2A=2-2^2+2^3-2^4+…+2^{101}-2^{101}` `⇒A+2A=1-2+2^2-2^3+…+2^100-2^101+(1-2+2^2-2^3+…+2^100-2^101)` `⇒3A=1-2^{101}` `⇒3A=1-2^{101}÷(1-2^{101})` `⇒3A=1` `⇒A=\frac{1}{3}` Bình luận
Cách giải: $A=(1-2+2^2-2^3+…+2^{100}-2^{101}):\underbrace{1-2^{102}}_{B}$ $→2A=2-2^2+2^3-2^4+….+2^{101}-2^{102}$ $→2A+A=3A=1-2^{102}$ $→A=\dfrac{1-2^{102}}{3}$ $→A:B=\dfrac{1-2^{102}}{3}:(1-2^{102})$ $=\dfrac{1}{3}$ Bình luận
Tham khảo
`A=1-2+2^2-2^3+…+2^{100}-2^{101}`
`⇒2A=2-2^2+2^3-2^4+…+2^{101}-2^{101}`
`⇒A+2A=1-2+2^2-2^3+…+2^100-2^101+(1-2+2^2-2^3+…+2^100-2^101)`
`⇒3A=1-2^{101}`
`⇒3A=1-2^{101}÷(1-2^{101})`
`⇒3A=1`
`⇒A=\frac{1}{3}`
Cách giải:
$A=(1-2+2^2-2^3+…+2^{100}-2^{101}):\underbrace{1-2^{102}}_{B}$
$→2A=2-2^2+2^3-2^4+….+2^{101}-2^{102}$
$→2A+A=3A=1-2^{102}$
$→A=\dfrac{1-2^{102}}{3}$
$→A:B=\dfrac{1-2^{102}}{3}:(1-2^{102})$
$=\dfrac{1}{3}$