Enunț
Problema rezolvată #16: Să se arate că numărul T este pătrat perfect.
T= \underbrace{111…1}_\text{n-1\;cifre} \underbrace{222…2}_\text{n\;cifre}5Rezolvare
Scriem numărul 1111…11 sub formă de sumă de puteri ale lui 10.
\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= \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) ^2În final, obținem deci că T poate fi scris sub forma unui pătrat perfect astfel:
T = \left( \frac{10^{n}+5}{3} \right) ^2[the_ad_group id=”102″]
