Question

Chứng minh tổng sau chia hết cho 5: 
S=3+3^2+3^3+...+3^100 

Answer 1

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

\(=3.\left(1+3+3^2+3^3\right)+...+3^{97}.\left(1+3+3^2+3^3\right)\)

\(=3.\left(1+3+9+27\right)+...+3^{97}.\left(1+3+9+27\right)\)

\(=3.40+...+3^{97}.40\)

\(=40.\left(3+...+3^{97}\right)\)

\(=5.8.\left(3+...+3^{97}\right)\text{chia hết cho 5}\)

=> S chia hết cho 5 =>đpcm.

Answer 2

S=3+3^2+3^3+....+3^100

S=(3+3^2+3^3+3^4)+....+(3^97+3^98+3^99+3^100)

S=1(3+3^2+3^3+3^4)+...+3^96.(3+3^2+3^3+3^4)

S=1.120+...+3^96.120

S=120(1+...+2^96)

S=5.24(1+...+2^96) chia hết cho 5