Question

Determinati n∈Ν astfel incat: $S= \frac{1}{ \sqrt{3+ 2\sqrt{2} } } + \frac{1}{ \sqrt{5+ 2\sqrt{6} } } +...+ \frac{1}{ \sqrt{2n+1+2 \sqrt{n(n+1)} } }$

Answer 1

$S=\dfrac{1}{\sqrt{(\sqrt2+\sqrt1)^2}}+\dfrac{1}{\sqrt{(\sqrt3+\sqrt2)^2}}+\dfrac{1}{\sqrt{(\sqrt4+\sqrt3)^2}}+...+\dfrac{1}{\sqrt{(\sqrt{n+1}+\sqrt n})^2}=$

$=\dfrac{1}{\sqrt2+\sqrt1}+\dfrac{1}{\sqrt3+\sqrt2}+\dfrac{1}{\sqrt4+\sqrt3}+...+\dfrac{1}{\sqrt{n+1}+\sqrt n}=$

$=\dfrac{\sqrt2-\sqrt1}{2-1}+\dfrac{\sqrt3-\sqrt2}{3-2}+\dfrac{\sqrt4-\sqrt3}{4-3}+...+\dfrac{\sqrt{n+1}-\sqrt n}{n+1-1}=$

$=\sqrt{n+1}-1$