Question

a) 3-3²+3³-3⁴+...+3⁹⁵-3⁹⁶

b) 1-2+2²-2³+...+2¹⁰⁰

c) (100-1²)x(100-2²)x(100-3²)x...x(100-99²)

Answer 1

a)Đặt \(A=3-3^2+3^3-3^4+...+3^{95}-3^{96}\)

\(3A=3^2-3^3+3^4-3^5+...+3^{96}-3^{97}\)

\(3A+A=\left(3^2-3^3+3^4-3^5+...+3^{96}-3^{97}\right)+\left(3-3^2+3^3-3^4+...+3^{95}-3^{96}\right)\)

\(4A=-3^{97}+3\)

\(A=\frac{-3^{97}+3}{4}\)

b)tương tự như câu a

c)\(\left(100-1^2\right)\left(100-2^2\right)\left(100-3^2\right).....\left(100-99^2\right)\)

\(=\left(10^2-1^2\right)\left(10^2-2^2\right)\left(10^2-3^2\right)....\left(10^2-10^2\right)...\left(10^2-99^2\right)\)

\(=\left(10^2-1^2\right)\left(10^2-2^2\right)\left(10^2-3^2\right)...0...\left(10^2-99^2\right)\)

=0

Answer 2

muốn chịch ko