Question

N=1+3^2+3^4+3^6+....+3^100

nhanh tay tìm N để có tích

Answer 1

\(N=1+3^2+3^4+...+3^{100}\)

\(\Rightarrow3^2N=3^2+3^4+...+3^{102}\)

\(\Rightarrow3^2N-N=\left(3^2+3^4+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)

\(\Rightarrow8N=3^{102}-1\)

\(\Rightarrow N=\frac{3^{102}-1}{8}\)

Answer 2

3²N=3²+3⁴+3⁶+3⁸+...+3¹⁰⁰+3¹⁰²=N+3¹⁰²

<=> 9N=N+3¹⁰²

=> 8N=3¹⁰²

=>\(N=\frac{3^{102}}{8}\)

Answer 3

N =1+ 3^{[(100 - 2 ) : 2 + 1 x (100 + 2 )] : 2} 

N = 1 + 3²⁵⁵⁰

N =  3⁰ + 3²⁵⁵⁰

N = 3²⁵⁵⁰