Question

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

Answer 1

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

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

=>\(4D-D=4^{101}+4^{100}+4^{99}+...+4^3+4^2-4^{100}-4^{99}-...-4^2-4\)

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

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

Answer 2

`#3107.101107`

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

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

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

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

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

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

Vậy, \(D=\dfrac{4^{101} - 4}{3}.\)