Question

Tính tổng S = 1 - 3 + 3² - 3³ + 3⁴ - .... + 3⁹⁸ - 3⁹⁹

Answer 1

Ta có: \(3S=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}\)

\(\Rightarrow4S=\left(1-3+3^2-3^3+3^4-...+3^{98}-3^{99}\right)+\left(3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}\right)\)

\(\Rightarrow4S=1-3^{100}\)

\(\Rightarrow S=\frac{1-3^{100}}{4}\)

 

Answer 2

 

S = 1 - 3 + 3² - 3³ + 3⁴ - .... + 3⁹⁸ - 3⁹⁹

 =1 -3.(1-3+3²-3³+3⁴-...+3⁹⁸-3⁹⁹+3⁹⁹)

=1-3(1 - 3 + 3²-3³+3⁴-...+3⁹⁸-3⁹⁹)-3.3⁹⁹

=1-3S-3¹⁰⁰

=>S+3S=1-3¹⁰⁰

=>4S=1-3¹⁰⁰

=>S=(1-3¹⁰⁰)/4