Question

Calculati:
a)$\sqrt{1+3+5+...+79}$
b)$\sqrt{1+3+5...+(2n-1)}$
c)$\sqrt{31+(2+4+...+60)}$
d)$\sqrt{(n+1)+(2+4+...+2n)}$

Answer 1

[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]