Question

cho B= 3 + 3² + 3³ + ... + 3¹⁰⁰. Tìm số tự nhiên n, biết 2.B + 3 = 3ⁿ

Answer 1

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

=>2B=3^101-3

=>2B+3=3^101

=>n=101

Answer 2

Ta có:\(B=3+3^2+3^3+...+3^{100}< =>3.B=3^2+3^3+3^4+....+3^{101}< =>3.B-B=2.B=\left(3^2+3^3+3^4+....+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)< =>2.B=3^{101}-3\)

Vậy 2.B+3=3¹⁰¹-3+3=3¹⁰¹ mà 2.B+3=3ⁿ<=>3ⁿ=3¹⁰¹<=>n=101

Vậy n=101

Answer 3

⇒3B=3²+3³+3⁴+....+3¹⁰¹

⇒2B=3¹⁰¹-3

⇒2B+3=3¹⁰¹-3+3

⇒2B=3¹⁰¹

⇒n=101

Answer 4

n=101