ta có :
1+3+32+33+34+....+350
\(A=1+3+3^2+....+3^{50}\)
\(3A=3+3^2+3^3....+3^{51}\)
\(3A-A=\left(3+3^2+3^3....+3^{51}\right)-\left(1+3^2++3^3+.....+3^{50}\right)\)
\(2A=3^{51}-1\)
\(A=\left(3^{51}-1\right):2\)
\(\Rightarrow\)1+3+32+33+34+....+350=(351-1):2