\(S=1+3+3^2+...+3^{99}\\ 3A=3+3^2+3^3+...+3^{100}\\ 3A-A=\left(3+3^2+3^3+...+3^{100}\right)-\left(1+3+3^2+...+3^{99}\right)\\ 2A=3^{100}-1\\ A=\dfrac{3^{100}-1}{2}\)
để ý rằng hầu hết các cơ số ở đây là 3 nên :
\(=>3S=3.\left(1+3^1+3^2+3^3+......+3^{99}\right)\)
\(=>3S=3+3^2+3^3+3^4+.....+3^{99}+3^{100}\)
\(=>3S-S=3^{100}-1\)
\(=>S=\dfrac{3^{100}-1}{2}\)
CHÚC BẠN HỌC TỐT.......
\(S=1+3+3^2+3^3+....+3^{99}\)
\(\Rightarrow3S=3\left(1+3+3^2+3^3+.....+3^{99}\right)\)
\(\Rightarrow3S=3+3^2+3^3+3^4+.....+3^{100}\)
\(\Rightarrow3S-S=\left(3+3^2+3^3+3^4+.....+3^{100}\right)-\left(1+3+3^2+3^3+.....+3^{99}\right)\)\(\Rightarrow2S=3^{100}-1\Rightarrow S=\dfrac{3^{100}-1}{2}\)
