Toán Lớp 6: Tìm số tự nhiên n để 2n – 1 – 2 – 2 ²- 2 ³-…- 2 mũ 100 =1
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Toán Lớp 6: Tìm số tự nhiên n để 2n – 1 – 2 – 2 ²- 2 ³-…- 2 mũ 100 =1
Comments ( 2 )
$\text{Ta có:2n-1-3-2^2-2^3-………-2^100=1}$
$\text{2n-(1+2^2+2^3+………+2^100)=1}$
$\text{ta đặt :}$
$\text{A=1+2^2+…+2^100}$
$\text{2A=2+2^2+…+2^101}$
$\text{2A-A=(2+2^2+…+3^101)-(1+2^2+…+2^100)}$
$\text{A=2^101-1}$
$\text{⇒2^n-(2^101-1)=1}$
$\text{⇒2^n-2^101+1=1}$
$\text{⇒2^n=1-1+2^101}$
$\text{⇒2^n=2^101}$
$\text{⇒n=101}$
$\text{Vậy n =101}$
$@Lun$
2^n – 1 – 2² – 2³ – … – 2¹⁰⁰ = 1
2^n = 1 + 1 +2² + 2³ + … + 2¹⁰⁰
2^n = 2 + 2² + 2³ + … + 2¹⁰⁰
2 . 2^n = 2.(2 + 2² + 2³ + …+ 2¹⁰⁰)
2.2^n = 2² + 2³ + 2⁴ + … + 2¹⁰¹
2 . 2^n – 2^n = (2² + 2³ + 2⁴ + … + 2¹⁰¹) – (2 + 2² + 2³ + … + 2¹⁰⁰)
2^n = 2¹⁰¹ – 2
Xin Hay Nhất Ạ+Sao=Cảm Ơn ạ