Tìm A biết: A = $\frac{1+2+2^2+2^3+2^4+…+2^{2019}}{1-2^{2020}}$ 18/09/2021 Bởi Peyton Tìm A biết: A = $\frac{1+2+2^2+2^3+2^4+…+2^{2019}}{1-2^{2020}}$
Đặt $B=1+2+2^2+2^3+2^4+…+2^{2019}$ $⇒2B=2+2^2+2^3+2^4+2^5+…+2^{2020}$ $⇒2B-B=2+2^2+2^3+2^4+2^5+…+2^{2020}-1-2-2^2-2^3-2^4-…-2^{2019}$ $⇔B=2^{2020}-1$ Khi đó $A=\dfrac{B}{1-2^{2020}}=\dfrac{2^{2020}-1}{1-2^{2020}}=\dfrac{-(1-2^{2020})}{1-2^{2020}}=-1$ Bình luận
Gọi B = 1 + 2^2 + 2^3 + 2^4 + …. + 2^2019 Ta có : 2B = 2 + 2^3 + 2^4 + 2^5+… + 2^2020 2B – B = 2 + 2^3 + 2^4 + 2^5+… + 2^2020 – 1 – 2^2 – 2^3 – 2^4 – …. – 2^2019 B = 1 – 2^2020 ⇒ A =$\frac{1 – 2^2020}{1- 2^2020}$ = 1 Vậy A = 1 Bình luận
Đặt $B=1+2+2^2+2^3+2^4+…+2^{2019}$
$⇒2B=2+2^2+2^3+2^4+2^5+…+2^{2020}$
$⇒2B-B=2+2^2+2^3+2^4+2^5+…+2^{2020}-1-2-2^2-2^3-2^4-…-2^{2019}$
$⇔B=2^{2020}-1$
Khi đó $A=\dfrac{B}{1-2^{2020}}=\dfrac{2^{2020}-1}{1-2^{2020}}=\dfrac{-(1-2^{2020})}{1-2^{2020}}=-1$
Gọi B = 1 + 2^2 + 2^3 + 2^4 + …. + 2^2019
Ta có : 2B = 2 + 2^3 + 2^4 + 2^5+… + 2^2020
2B – B = 2 + 2^3 + 2^4 + 2^5+… + 2^2020 – 1 – 2^2 – 2^3 – 2^4 – …. – 2^2019
B = 1 – 2^2020
⇒ A =$\frac{1 – 2^2020}{1- 2^2020}$ = 1
Vậy A = 1