[tex]a)\sqrt{1+3+5+...+79}=\\
\sqrt{1+2+3+....+79-2-4-6-...-78}=\\
\sqrt{\frac{79*80}{2}-2(1+2+3+39})=\\
\sqrt{79*40-2*\frac{39*40}{2}}=\\
\sqrt{79*40-39*40}=\\
\sqrt{40(79-39)}=\\
\sqrt{40^2}=40\\
b)\sqrt{1+3+5+....+(2n-1)}=\\
\sqrt{1+2+3+....+(2n-1)-2-4-....-(2n-2)}=\\
\sqrt{\frac{(2n-1)*2n}{2}-2[1+2+3+...+(n-1)]}=\\
\sqrt{(2n-1)*n-2*\frac{(n-1)*n}{2}}=\\
\sqrt{(2n-1)*n-(n-1)*n}=\\
\sqrt{n(2n-1-n+1)}=\\
\sqrt{n^2}=n\\
[/tex]
[tex]c)\sqrt{31+2+4+6+....+60}=\\
\sqrt{31+2(1+2+3+....+30)}=\\
\sqrt{31+2*\frac{30*31}{2}}=\\
\sqrt{31+30*31}=\\
\sqrt{31(1+30)}=\\
\sqrt{31^2}=31\\
d)\sqrt{(n+1)+(2+4+6+....+2n)}=\\
\sqrt{(n+1)+2(1+2+3+.....+n)}=\\
\sqrt{(n+1)+2*\frac{n(n+1)}{2}}=\\
\sqrt{(n+1)+n(n+1)}=\\
\sqrt{(n+1)(n+1)}=\\
\sqrt{(n+1)^2}=n+1[/tex]
