Question

Cho \(S=2+2^2+2^4+2^6+...+2^{102}\)

a) Tính gọn S

b) Tìm số dư khi chia S cho 3

Answer 1

Ta có :

\(S=2^0+2^2+2^4+............+2^{102}\)

\(\Leftrightarrow2^2S=2^2+2^4+..............+2^{102}+2^{104}\)

\(\Leftrightarrow4S-S=\left(2^2+2^4+.........+2^{104}\right)-\left(1+2^2+.........+2^{102}\right)\)

\(\Leftrightarrow3S=2^{104}-1\)

\(\Leftrightarrow S=\dfrac{2^{104}-2}{3}\)

Answer 2

a) \(S=2+2^2+2^4+2^6+...+2^{102}\)

\(\Rightarrow2^2S=2^2\left(2+2^2+2^4+2^6+...+2^{102}\right)\)

\(\Leftrightarrow4S=2^3+2^4+2^6+2^8+...+2^{104}\)

\(\Rightarrow4S-S=3S=\left(2^3+2^4+2^6+2^8+...+2^{104}\right)-\left(2+2^2+2^4+2^6+...+2^{102}\right)\)

\(3S=2^3+2^{104}-2-2^2\)

\(3S=8+2^{104}-2-4=2^{104}-2\)

\(\Rightarrow S=\dfrac{2^{104}-2}{3}=\)