\(A=1+3^2+3^4+3^6+.....+3^{100}\)
\(A=1+3^2+3^4+3^6+3^8+.......+3^{98}+3^{100}\)( A có 51 số hạng )
\(A=1+\left(3^2+3^4\right)+\left(3^6+3^8\right)+.....+\left(3^{98}+3^{100}\right)\)
\(A=1+3^2\left(1+3^2\right)+3^6\left(1+3^2\right)+.....+3^{98}+\left(1+3^2\right)\)
\(A=1+3^2\left(1+9\right)+3^6\left(1+9\right)+.....+3^{98}\left(1+9\right)\)
\(A=1+3^2.10+3^6.10+.....+3^{98}.10\)
\(A=1+10\left(3^2+3^6+.....+3^{98}\right)\)
Vì \(10⋮10\Rightarrow10.\left(3^2+3^6+....+3^{98}\right)⋮10\)
Mà \(1:10\)dư \(1\)
\(\Rightarrow1+10\left(3^2+3^6+....+3^{98}\right):10\)dư \(1\)
Vậy A chia 10 dư 1