Question

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

Answer 1

Lời giải:

Ta có \(A=4+4^2+4^3+....+4^{100}\)

\(\Rightarrow 4A=4^2+4^3+...+4^{100}+4^{101}\)

Trừ theo vế:

\(\Rightarrow 3A=4^{101}-4\Rightarrow A=\frac{4^{101}-4}{3}\)

Answer 2

Đặt:
\(S=4+4^2+4^3+4^4+...+4^{100}\)

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

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

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

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