1) Avem [tex]\displaystyle\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}[/tex].
Dând lui k valori de la 1 la n avem
[tex]\displaystyle\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\ldots+\frac{1}{n(n+1}=\\=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{n}-\frac{1}{n+1}=\\=1-\frac{1}{n+1}=\frac{n}{n+1}[/tex]
Rezultă n=100.
2) Suma se scrie
[tex]\left(\displaystyle\frac{12}{11}-1\right)+\left(\displaystyle\frac{13}{22}-\frac{1}{2}\right)+\ldots+\left(\displaystyle\frac{110}{1089}-\frac{1}{99}\right)=\\=\displaystyle\frac{1}{2}+\frac{2}{22}+\frac{3}{33}+\ldots+\frac{99}{1089}=\\=\underbrace{\displaystyle\frac{1}{11}+\frac{1}{11}+\ldots+\frac{1}{11}}_{\mbox{99 ori}}=\frac{99}{11}=9[/tex]
Deci suma este pătratul lui 3.
3) Ecuația se scrie
[tex]\displaystyle\frac{x}{99}+\frac{10+x}{99}+\frac{20+x}{99}+\ldots+\frac{80+x}{99}=x[/tex]
sau [tex]9x+10+20+\ldots+80=99x\Rightarrow 90x=360\Rightarrow x=4[/tex]
4) [tex]\sqrt{3^4+3^5}=\sqrt{3^4(1+3)}=\sqrt{81\cdot 4}=9\cdot 2=18[/tex]
