Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 163957 by HongKing last updated on 12/Jan/22

Answered by mahdipoor last updated on 12/Jan/22

$$getn=3b+a,b\in N0⩽a<3$$$$\left[n\right]=3b+\left[a\right]\left[\frac{n+1}{3}\right]=b+\left[\frac{a+1}{3}\right]$$$$\left[\frac{n}{3}\right]=b+\left[\frac{a}{3}\right]=b\left[\frac{n+2}{3}\right]=b+\left[\frac{a+2}{3}\right]$$$$\Rightarrow {(-1)}^{3b+\left[a\right]}+{(-1)}^{b+\left[\frac{a+1}{3}\right]}={(-1)}^b+{(-1)}^{b+\left[\frac{a+2}{3}\right]}$$$$\Rightarrow {(-1)}^b({(-1)}^{\left[a\right]}+{(-1)}^{\left[\frac{a+1}{3}\right]})={(-1)}^b(1+{(-1)}^{\left[\frac{a+2}{3}\right]})$$$$\Rightarrow {(-1)}^{\left[a\right]}+{(-1)}^{\left[\frac{a+1}{3}\right]}=1+{(-1)}^{\left[\frac{a+2}{3}\right]}$$$$\Rightarrow \{\begin{array}{l}0⩽a<1\Rightarrow 1+1=1+1\\ 1⩽a<2\Rightarrow -1+1=1-1\\ 2⩽a<3\Rightarrow 1-1=1-1\end{array}$$

Commented by HongKing last updated on 13/Jan/22

$$\mathrm{cool}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com