Question

Tính tổng D= 4 + 4^2 + 4^3 + 4^4 + … + 4^100

Answer 1

\(D=4+4^2+4^3+4^4+...+4^{100}\)

\(4D=4^2+4^3+4^4+4^5+...+4^{101}\)

\(4D-D=\left(4^2+4^3+4^4+4^5+...+4^{101}\right)-\left(4+4^2+4^3+4^4+...+4^{100}\right)\)

\(3D=4^{101}-4\)

\(D=\dfrac{4^{101}-4}{3}\)

\(#WendyDang\)

Answer 2

D=1,606938044*10^60

Answer 3

D = 4 + 4² + 4³ + ... + 4¹⁰⁰

⇒ 4D = 4² + 4³ + 4⁴ + ... + 4¹⁰¹

⇒ 3D = 4D - D

= (4² + 4³ + 4⁴ + ... + 4¹⁰¹) - (4 + 4² + 4³ + ... + 4¹⁰⁰)

= 4¹⁰¹ - 4

⇒ D = (4¹⁰¹ - 4) : 3

Answer 4

\(D=4+4^2+4^3+4^4+...+4^{100}\\ \Rightarrow4D=4^2+4^3+4^4+4^5+...+4^{101}\\\Rightarrow 4D-D=4^2+4^3+4^4+4^5+...+4^{101}-4+4^2+4^3+4^4+...+4^{100}\\ \Rightarrow3D=\left(4^2-4^2\right)+\left(4^3-4^3\right)+...+\left(4^{100}-4^{100}\right)+4^{101}-4\\ \Rightarrow3D=4^{101}-4\\ \Rightarrow D=\dfrac{4^{101}-4}{3}\)

Vậy \(D=\dfrac{4^{101}-4}{3}\)