Question

va rog mult o sa dau coroana macar unul din ele sa il faceti

Answer 1

Răspuns

Explicație pas cu pas:

$3. a)1+2+2^2+\ldots+2^{2018}=2^x-1\\ \text{Notam suma cu S}.\\S=1+2+2^2+\ldots+2^{2018}\\2S=2+2^2+2^3+\ldots+2^{2019}\\\text{Daca scadem a doua relatie din prima se obtine}~ S=2^{2019}-1\\Ecuatia~devine~2^{2019}-1=2^x-1 ,de~unde~\boxed{\bold{x=2019}}\\b)\text{Se calculeaza ciclul puterilor lui 2.}\\u(2^1)=2~~~u(2^2)=4~~~u(2^3)=8~~~u(2^4)=6\\2019:4=504,rest=3 ,ceea~ce~inseamna~ca~u(2^{2019})=8\\Prin~urmare~u(2^{2019}-1)=u(2^{2019})-1=8-1=\boxed{\bold{7}}$

$a)\overline{1abcd}+\overline{abcd5}=47438\\10000+\overline{abcd}+10\cdot\overline{abcd}+5=47438\\11\cdot\overline{abcd}+10005=47438\\11\cdot \overline{abcd}=37433\\\boxed{\bold{\overline{abcd}=3403}}$

$b)1+3+5+\ldots+49=\\1+2+3+\ldots+49-2-4-\ldots-48=\\\\\dfrac{49\cdot50}{2}-2(1+2+3+\ldots+24)=\\49\cdot25-\not2\cdot \dfrac{24\cdot 25}{\not 2}=\\49\cdot25-24\cdot25=\\25(49-24)=25\cdot 25=25^2=patrat~perfect$