Question

Demonstrati ca urmatoarele numere naturale sunt patrate perfecte:
[tex]~\underbrace {999...}~\underbrace{94000...09}\\
~~~ ~n~ cifre ~~~~~n~cifre [/tex]

Answer 1

Salut,

Enunțul corect este: demonstrați că numărul N este pătrat perfect, unde n≥2:

$N=\underbrace{99...9}_{de\;n-1\;ori}94\underbrace{00...0}_{de\;n\;ori}9.\\\\N=\underbrace{99...9}_{de\;n-1\;ori}\underbrace{00...0}_{de\ n+3\;ori}+94\underbrace{00...0}_{de\ n\ ori}9\\\\Pentru\;primul\;num\breve{a}r:\\900000=10^6-10^5,\;n=2\ (6=num\breve{a}rul\;total\;de\;cifre,\;iar\;5=num\breve{a}rul\;de\;zerouri);\\99000000=10^8-10^6,\;n=3\ (8=num\breve{a}rul\;total\;de\;cifre,\;iar\;6=num\breve{a}rul\;de\;zerouri);\\...\\\underbrace{99...9}_{de\;n-1\;ori}\underbrace{00...0}_{de\ n+3\;ori}=10^{2n+2}-10^{n+3}.\\\\Pentru\;al\;doilea\;num\breve{a}r:\\94009=94\cdot10^3+9,\;n=2\ (3=num\breve{a}rul\;de\;zerouri+1);\\940009=94\cdot10^4+9,\;n=3\ (4=num\breve{a}rul\;de\;zerouri+1);\\...\\94\underbrace{00...0}_{de\ n\ ori}9=94\cdot10^{n+1}+9.\\N=10^{2n+2}-10^{n+3}+94\cdot10^{n+1}+9=10^{2n+2}-10^{n+3}+(100-6)\cdot10^{n+1}+9=10^{2n+2}-10^{n+3}+10^{n+3}-6\cdot10^{n+1}+9=(10^{n+1}-3)^2,\;care\;este\;p\breve{a}trat\;perfect.\\Green\;eyes.$