Question

F=2+2^2+2^3+...+2^100

E=3^0+3^1+3^2+...+3^100

giúp mik với nhé mik cảm ơn!!!!!<3

Answer 1

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

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

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

\(F=2^{101}-2\)

Vậy...

\(E=3^0+3^1+3^2+...+3^{100}\)

\(E=1+3+3^2+...+3^{100}\)

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

\(3E-3E=\left(3+3^2+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)

\(2E=3^{101}-1\)

\(E=\frac{3^{101}-1}{2}\)

Vậy...