[jacks]

If \( \displaystyle S =\sum^{99}_{r=1}\frac{1}{\sqrt{r}+\sqrt{r+1}}. \) Then \( \displaystyle S+\sqrt{3}-\sqrt{2}= \)

[Luis Fuentes]

Hi

If \( \displaystyle S =\sum^{99}_{r=1}\frac{1}{\sqrt{r}+\sqrt{r+1}}. \) Then \( \displaystyle S+\sqrt{3}-\sqrt{2}= \)

Hint:

\( \dfrac{1}{\sqrt{r}+\sqrt{r+1}}=\dfrac{\sqrt{r+1}-\sqrt{r}}{(\sqrt{r+1}+\sqrt{r})(\sqrt{r+1}-\sqrt{r})}=\dfrac{\sqrt{r+1}-\sqrt{r}}{(r+1)-r}=\sqrt{r+1}-\sqrt{r} \)

Best regards.

[jacks]

Thanks Admin.

Actually question as \( \displaystyle S=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\cdots \cdots +\frac{1}{\sqrt{99}+\sqrt{100}}. \) Then \( \displaystyle S +\sqrt{3}-\sqrt{2} \)

[Luis Fuentes]

Hi

Thanks Admin.

Actually question as \( \displaystyle S=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\cdots \cdots +\frac{1}{\sqrt{99}+\sqrt{100}}. \) Then \( \displaystyle S +\sqrt{3}-\sqrt{2} \)

Sure that this is the right statment? I think that this sum cannot be computed explictly. It's equivalent to:

\( \displaystyle\sum_{n=1}^{50}{}\sqrt{2n}-\sqrt{2n-1} \)

Best regards.

[jacks]

Thanka so much Admin.