Problema rezolvată #16 – 111…1222…25 pătrat perfect

Enunț

Problema rezolvată #16: Să se arate că numărul T este pătrat perfect.

T=1111n-1  cifre2222n  cifre5 T= \underbrace{111…1}_\text{n-1\;cifre} \underbrace{222…2}_\text{n\;cifre}5
Rezolvare

Scriem numărul 1111…11 sub formă de sumă de puteri ale lui 10.

1111n=100+101+...+10n1=  =10n19\underbrace{111…1}_\text{n}= 10^0+10^1+...+10^{n-1}=\\\;\\=\frac{10^n-1}{9}

Atunci, numărul T se poate scrie astfel:

T=11n-110n+1+22n10+5=  =11n-110n+1+211n10+5=  =10n11910n+1+   +210n1910+5=  =19{(10n11)10n+1+   +2(10n1)10+45}=  =19(10n110n+110n+1+   +210n10210+45)=  =19(102n+10n+1+25)=  =19(102n+1010n+25)=  =19(102n+2510n+52)=  =19(10n+5)2=  =(10n+53)2T= \underbrace{1…1}_\text{n-1} \cdot10^{n+1}+\underbrace{2…2}_\text{n}\cdot 10 +5=\\\;\\= \underbrace{1…1}_\text{n-1} \cdot10^{n+1}+ 2\cdot\underbrace{1…1}_\text{n}\cdot 10 +5 =\\\;\\= \frac{10^{n-1}-1}{9}\cdot10^{n+1}+\\\;\\\ +2\cdot\frac{10^{n}-1}{9}\cdot10+5=\\\;\\= \frac{1}{9}\{(10^{n-1}-1)\cdot10^{n+1}+\\\;\\\ +2\cdot(10^{n}-1)\cdot10+45\}=\\\;\\= \frac{1}{9}(10^{n-1}\cdot10^{n+1}-10^{n+1}+\\\;\\\ +2\cdot10^{n}\cdot10-2\cdot10 +45)=\\\;\\= \frac{1}{9}(10^{2n}+10^{n+1}+25)=\\\;\\= \frac{1}{9}(10^{2n}+10\cdot10^{n}+25)=\\\;\\= \frac{1}{9}(10^{2n}+2\cdot5\cdot10^{n}+5^2)=\\\;\\= \frac{1}{9}(10^{n}+5)^2=\\\;\\= \left( \frac{10^{n}+5}{3} \right) ^2

În final, obținem deci că T poate fi scris sub forma unui pătrat perfect astfel:

T=(10n+53)2T = \left( \frac{10^{n}+5}{3} \right) ^2

[the_ad_group id=”102″]

Pe aceeași temă