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Question Number 83931 by john santu last updated on 08/Mar/20

$$\frac{1}{(\sqrt{1}+\sqrt{2})(\sqrt[4]{1}+\sqrt[4]{2})}+\frac{1}{(\sqrt{2}+\sqrt{3})(\sqrt[4]{2}+\sqrt[4]{3})}+$$$$\frac{1}{(\sqrt{3}+\sqrt{4})(\sqrt[4]{3}+\sqrt[4]{4})}+...+\frac{1}{(\sqrt{255}+\sqrt{256})(\sqrt[4]{255}+\sqrt[4]{256})}$$$$=...$$

Commented by john santu last updated on 08/Mar/20

$$\mathrm{yes}.\mathrm{the}\mathrm{answer}3$$

Answered by TANMAY PANACEA last updated on 08/Mar/20

$$T_r=\frac{1}{(\sqrt{r}+\sqrt{r+1})(\sqrt[4]{r}+\sqrt[4]{r+1)}}$$$$S=\underset{r=1}{\sum ^{255}}{T}_r$$$$T_r=\frac{(r+1)-r}{D_r}=\frac{\sqrt{r+1}-\sqrt{r}}{(\sqrt[4]{r+1}+\sqrt[4]{r})}=\frac{(\sqrt[4]{r+1}+\sqrt[4]{r})\times (\sqrt[4]{r+1}-\sqrt[4]{r})}{(\sqrt[4]{r+1}+\sqrt[4]{r})}$$$$T_r=\sqrt[4]{r+1}-\sqrt[4]{r}$$$$so\underset{r=1}{\sum ^{255}}{T}_r=0\mathit{p}\mathit{l}\mathit{s}\mathit{c}\mathit{h}\mathit{e}\mathit{c}\mathit{k}\leftarrow itismyerror$$$$soplsignore...$$$$editing...$$$$T_r=\sqrt[4]{r+1}-\sqrt[4]{r}$$$$T_1=\sqrt[4]{2}-\sqrt[4]{1}$$$$T_2=\sqrt[4]{3}-\sqrt[4]{2}$$$$...$$$$...$$$$T_{255}=\sqrt[4]{256}-\sqrt[4]{255}$$$$S=\sqrt[4]{256}-\sqrt[4]{1}=4-1=3$$

Commented by JDamian last updated on 08/Mar/20

$$ifyourexpressionfor{T}_{r}iscorrect,then$$$$S{=}^4\sqrt{255+1}{-}^4\sqrt{1}=4-1=3$$

Commented by jagoll last updated on 08/Mar/20

$$\mathrm{why}\underset{\mathrm{r}=1}{\sum ^{255}}{\mathrm{T}}_{\mathrm{r}}=\underset{\mathrm{r}=1}{\sum ^{255}}(\sqrt[4]{\mathrm{r}+1}-\sqrt[4]{\mathrm{r}})=0?$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{typo}\mathrm{sir}?$$

Commented by TANMAY PANACEA last updated on 08/Mar/20

$$yeshurry$$

Commented by john santu last updated on 08/Mar/20

$$\mathrm{thank}\mathrm{you}$$

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