Question

Cho S = 1+3²+3⁴+3⁶+.....................................+3⁹⁸.Tính tổng của S và chứng minh S chia hết cho 10

giúp mik nha

Answer 1

S=1+3²+3⁴+3⁶+.............................+3⁹⁸

9S=3+3⁴+3⁶+3⁸+.........................+3¹⁰⁰

=> 9S-S=3¹⁰⁰-1

3¹⁰⁰-1=(3⁴)²⁵-1

=(...1)²⁵-1

=(.....1)-1

=(.....0) chia hết cho 10

Vậy S chia hết cho 10

Answer 2

a, \(S=1+3^2+3^4+3^6+...+3^{98}\)

\(\Rightarrow3^2S=3^2+3^4+3^6+3^8+...+3^{100}\)

\(\Rightarrow3^2S-S=\left(3^2+3^4+3^6+3^8+...+3^{100}\right)-\left(1+3^2+3^4+3^6+...+3^{98}\right)\)

\(\Rightarrow8S=3^{100}-1\)

\(\Rightarrow S=\frac{3^{100}-1}{8}\)

Vậy : \(S=\frac{3^{100}-1}{8}\)

b, \(S=1+3^2+3^4+3^6+...+3^{98}\)

\(S=\left(1+3^2\right)+\left(3^4+3^6\right)+...+\left(3^{96}+3^{98}\right)\)

\(S=\left(1+3^2\right)+3^4\left(1+3^2\right)+...+3^{96}\left(1+3^2\right)\)

\(S=1.10+3^4.10+...+3^{96}.10\)

\(S=\left(1+3^4+...+3^{96}\right).10\)

Vì : \(1+3^4+...+3^{96}\in N\Rightarrow S⋮10\)

Vậy : \(S⋮10\)