Question

Tính E=1+4+4^2+4^3+...+4^98+4^99

Answer 1

\(E=1+4+4^2+4^3+...+4^{98}+4^{99}\)

\(\Rightarrow4E=4+4^2+4^3+...+4^{99}+4^{100}\)

\(\Rightarrow4E-E=\left(4+4^2+4^3+...+4^{99}+4^{100}\right)-\left(1+4+4^2+4^3+...+4^{98}+4^{99}\right)\)

\(\Rightarrow3E=\left(4-4\right)+\left(4^2-4^2\right)+\left(4^3+4^3\right)+...+\left(4^{99}-4^{99}\right)+\left(4^{100}-1\right)\)

\(\Rightarrow3E=4^{100}-1\)

\(\Rightarrow E=\dfrac{4^{100}-1}{3}\)

Answer 2

4E=4+4^2+...+4^100

=>3E=4^100-1

=>E=(4^100-1)/3