Question

S = 1+3¹+3²+3³+...+3⁹⁹

Answer 1

\(S=1+3+3^2+...+3^{99}\\ 3A=3+3^2+3^3+...+3^{100}\\ 3A-A=\left(3+3^2+3^3+...+3^{100}\right)-\left(1+3+3^2+...+3^{99}\right)\\ 2A=3^{100}-1\\ A=\dfrac{3^{100}-1}{2}\)

Answer 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.......

Answer 3

\(S=1+3+3^2+3^3+....+3^{99}\)

\(\Rightarrow3S=3\left(1+3+3^2+3^3+.....+3^{99}\right)\)

\(\Rightarrow3S=3+3^2+3^3+3^4+.....+3^{100}\)

\(\Rightarrow3S-S=\left(3+3^2+3^3+3^4+.....+3^{100}\right)-\left(1+3+3^2+3^3+.....+3^{99}\right)\)\(\Rightarrow2S=3^{100}-1\Rightarrow S=\dfrac{3^{100}-1}{2}\)