1+2+2^2+2^3+2^4+2^5+……+2^2017 Tìm số tự nhiên n,biết 2.(A+1)=2^n+1 25/11/2021 Bởi Iris 1+2+2^2+2^3+2^4+2^5+……+2^2017 Tìm số tự nhiên n,biết 2.(A+1)=2^n+1
Đặt N = 1 + 2 + 2^2 + 2^3+2^4+2^5+….+2+2017 ⇒2N = 2 ( 1+2+2^2+2^3+2^4+2^5+……+2^2017) 2N= 2+ 2^2+2^3+2^4+2^5+2^6+……+2^2018 ⇒2N-N=(2+2^2+2^3+2^4+2^5+2^6+…….+2^2018) – ( 1+ 2+2^2+2^3+2^4+2^5+……..+2^2017) = 2^2018 – 1 Bình luận
Đặt N = 1 + 2 + 2^2 + 2^3+2^4+2^5+….+2+2017
⇒2N = 2 ( 1+2+2^2+2^3+2^4+2^5+……+2^2017)
2N= 2+ 2^2+2^3+2^4+2^5+2^6+……+2^2018
⇒2N-N=(2+2^2+2^3+2^4+2^5+2^6+…….+2^2018) – ( 1+ 2+2^2+2^3+2^4+2^5+……..+2^2017)
= 2^2018 – 1