Sir cu o infinitate de patrate perfecte(own)

Claudiu Mindrila · Post by **Claudiu Mindrila** » Wed Mar 25, 2009 8:26 pm

Aratati ca \( 1779552-5328, \: 177795552-53328, \: 17777955552-533328, \dots \) sunt patrate perfecte.

Claudiu Mindrila, Revista Minus 1/2009

Marius Mainea · Post by **Marius Mainea** » Fri Mar 27, 2009 9:52 pm

Termenul general al sirului este \( a_n=1\underbrace{77...7}_{n ori}4\underbrace{22...2}_{n ori}4=(4\cdot3\cdot\underbrace{11...1}_{n ori})^2 \)

Alte probleme de acest fel:

1) Sa se arate ca numarul \( A=\underbrace{44...4}_{2n ori}-\underbrace{88...8}_{nori} \) este patrat perfect. Care este acest patrat?

2) Sa se arate ca numarul \( B=\underbrace{11...1}_{n-1 ori}\underbrace{22...2}_{nori}5 \) este patrat perfect. Care este acest patrat?

3) Numarul \( C=\frac{1}{3}\underbrace{88...8}_{3nori}-\underbrace{88...8}_{nori}\cdot 10^n \) este cub perfect. Care este acest cub?

Post by **Laurian Filip** » Fri Mar 27, 2009 11:13 pm

Forma generala este de fapt \( a_n=1\underbrace{77...7}_{n ori}4\underbrace{22...2}_{n ori}4 \) . In plus patratul perfect pe care l-ati scris are ultima cifra 6, deci nu poate fi egal cu \( a_n \)

Marius Mainea · Post by **Marius Mainea** » Fri Mar 27, 2009 11:15 pm

Da , scuze.

Claudiu Mindrila · Post by **Claudiu Mindrila** » Sat Mar 28, 2009 11:35 am

Marius Mainea wrote:Termenul general al sirului este \( a_n=1\underbrace{77...7}_{n ori}4\underbrace{22...2}_{n ori}4=(4\cdot3\cdot\underbrace{11...1}_{n ori})^2 \)

Solutia mea: \( 1\underbrace{77\dots7}_{n-1\ ori}9\underbrace{55\dots5}_{n-1\ ori}2-5\underbrace{33\dots3}_{n-2\ ori}28=16\cdot\frac{10^{n}-1}{9}\cdot10^{n}+32\cdot\frac{10^{n}-1}{9}-\frac{16}{3}\left(10^{n}-1\right)=\frac{16}{9}\left(10^{n}-1\right)\left(10^{n}+2-3\right)=\left[\frac{4}{3}\left(10^{n}-1\right)\right]^{2} \)