A=1+3+32+...+399
3A=3+32+33+...+3100
3A+1=(1+3+32+...+399)+3100
3A+1=A+3100
3A-A=3100-1
2A=3100-1
A=(3100-1)/2
Ta có: A = 1 + 3 + 32 + .... + 399
=> 3A = 3 + 32 + 33 + .... + 3100
=> 3A - A = 3100 - 1
=> 2A = 3100 - 1
=> A = 3100 - 1/2
\(A=1+3+3^2+...+3^{99}\)
\(3A=3+3^2+3^3+...+3^{100}\)
\(3A-A=\left(3+3^2+3^3+...+3^{100}\right)-\left(1+3+3^2+...+3^{99}\right)\)
\(2A=3^{100}-1\)
\(A=\frac{3^{100}-1}{2}\)