\(B=3^1+3^2+3^3+3^4+.....+3^{99}+3^{100}\)
\(B=\left(3^1+3^2\right)+\left(3^3+3^4\right)+....+\left(3^{99}+3^{100}\right)\)
\(B=3.\left(3+1\right)+3^3.\left(3+1\right)+....+3^{99}.\left(3+1\right)\)
\(B=\left(3+1\right).\left(3+3^3+.........+3^{99}\right)\)
\(B=4.\left(3+3^3+..........+3^{99}\right)\)
Vì 4 chia hết cho 4 nên \(4.\left(3+3^3+..........+3^{99}\right)\) chia hết cho 4
Do đó \(3^1+3^2+3^3+3^4+.....+3^{99}+3^{100}\) chia hết cho 4
Vậy B chia hết cho 4
Chúc bạn học tốt!!!
\(B=3^1+3^2+3^3+3^4+.....+3^{100}\)
\(B=\left(3^1+3^2\right)+\left(3^3+3^4\right)+.....+\left(3^{99}+3^{100}\right)\)
\(B=3\left(1+3\right)+3^2\left(3+1\right)+.....+3^{98}\left(3+1\right)\)
\(B=3.4+3^2.4+.....3^{98}.4\)
\(B=4\left(3+3^2+.....+3^{99}\right)\)
\(B⋮4\)
\(\rightarrowđpcm\)
\(B=3+3^2+...+3^{100}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)\)
\(=\left(3+3^3+...+3^{99}\right).4⋮4\)
\(\Rightarrowđpcm\)
B = 31 + 32 + 33 + 34 + ... + 3100
=> B = (31 + 32) + (33 + 34) + ... + (399 + 3100)
=> B = 31. (1 + 31) + 33. (1 + 31) + ... + 399. (1 + 31)
=> B = 31. 4 + 33. 4 + ... + 399. 4
=> B = 4 . (31 + 33 + ... + 399) \(⋮\) 4
