Question

Aratati ca numarul $S=1+2*2+3*2 ^{2} +...+2017*2 ^{2016}$ nu este patrat perfect.
Dau coroana(40 puncte)

Answer 1

[tex]S=1+2\cdot 2+3\cdot 2^2+_{\dots}+2017\cdot 2^{2016}|\cdot 2\\ 2S=2+2\cdot 2^2+3\cdot 2^3+_{\dots}+2017\cdot 2^{2017}\\ ------------------\\ 2S-S=2017\cdot 2^{2017}-2^{2016}-2^{2015}-_{\dots}-2-1\\ S=2017\cdot 2^{2017}-(2^{2016}+2^{2015}+_{\dots}+ 1)\\ *Notam:\\ K=1+2+2^2+_{\dots}+2^{2016}|\cdot 2\\ 2K=2+2^2+_{\dots}+2^{2017}\\ -----------\\ 2K-K=2^{2017}-1\\ K=2^{2017}-1\\ Revenind:\\ S=2017\cdot 2^{2017}-2^{2017}+1\\ S=2016\cdot 2^{2017}+1\\ [/tex]
[tex]\text{Mai departe aflam ultima cifra a lui S.}\\ u(S)=u (2016)\cdot u(2^{2017})+1\\ u(S)=6\cdot u(2^{2017})+1\\ 2^1=2, 2^2=4, 2^3=8, 2^4=..6\\ 2017:4=504,r=1\Rightarrow u(2^{2017})=2\\ Asadar:\\ u(S)=6\cdot2+1\\ u(S)=3\\ \text{Un patrat perfect nu poate avea ultima cifra 2,3,7 sau 8.} q.e.d\\ \text{Observatie: Cu u(x) s-a notat ultima cifra a numarului x.}[/tex]