Question

cho a=3+3²+3³+....+3¹⁰⁰. tìm số tự nhiên n biết rằng 2a+3=3ⁿ

Answer 1

a=3+3²+3³+....+3¹⁰⁰

=>3a=3²+3³+....+3¹⁰¹

=>3a-a=3²+3³+....+3¹⁰¹ -(3+3²+3³+....+3¹⁰⁰)

=>2a=3²+3³+....+3¹⁰¹-3-3²-3³-...-3¹⁰⁰

=>2a=3¹⁰¹-3

=>2a+3=3¹⁰¹

mà theo đề 2a+3=3ⁿ

=>n=101

vậy n=101

Answer 2

a=3+3²+...+3¹⁰⁰

=>3a=3²+3³+...+3¹⁰¹=> 3a-a=2a=(3²+3³+...+3¹⁰¹)-(3+3²+...+3¹⁰⁰)=3¹⁰¹-3

\(\Rightarrow a=\frac{3^{101}-3}{2}\)

=>2a+3=\(2\times\frac{3^{101}-3}{2}+3=\left(3^{101}-3\right)+3=3^{101}-3+3=3^{101}-\left(3-3\right)=3^{101}-0=3^{101}\)

Answer 3

mk quên bổ sung thêm nè 

2a+3=3¹⁰¹=3ⁿ=>n=101

Answer 4

n = 101 nha bn

Answer 5

n = 101 nha bn

Answer 6

n = 101 nha ban

k tui nha

thanks

Answer 7

Vậy n = 101

Mình tính rùi

Answer 8

\(n=101\)

Answer 9

\(a=3+3^2+3^3+...+3^{100}\)

\(\Rightarrow3a=3^2+3^3+...+3^{101}\)

\(\Rightarrow3a-a=3^2+3^3+...+3^{101}-\left(3+3^2+3^3+...+3^{100}\right)\)

\(\Rightarrow2a=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\)

\(\Rightarrow2a=3^{101}-3\)

\(\Rightarrow2a+3=3^{101}\)

Mà theo đề bài \(2a+3=3^n\)

\(\Rightarrow n=101\)

Vậy \(n=101\)