Question

tính

\(B=xy^2z^3+x^2y^3z^4+x^3y^4z^5+...+x^{2017}y^{2018}z^{2019}vớix=y=z=-1\)

Answer 1

Lời giải:

Với \(x=y=z=-1\) thì:

\(B=(-1)(-1)^2(-1)^3+(-1)^2(-1)^3(-1)^4+(-1)^3(-1)^4(-1)^5+...+(-1)^{2017}(-1)^{2018}(-1)^{2019}\)

\(=(-1)^{1+2+3}+(-1)^{2+3+4}+(-1)^{3+4+5}+...+(-1)^{2017+2018+2019}\)

\(=(-1)^6+(-1)^{9}+(-1)^{12}+...+(-1)^{6054}\)

\(=[(-1)^6+(-1)^{12}+(-1)^{18}+...+(-1)^{6054}]+[(-1)^9+(-1)^{15}+...+(-1)^{6051}]\)

\(=\underbrace{1+1+..+1}_{1009}+\underbrace{[(-1)+(-1)+..+(-1)]}_{1008}\)

\(=1009-1008=1\)