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Question Number 103673 by bobhans last updated on 16/Jul/20

$$\underset{k=1}{\sum ^{4095}}\frac{1}{(\sqrt{k}+\sqrt{k+1})(\sqrt[4]{k}+\sqrt[4]{k+1})}?$$

Answered by bramlex last updated on 16/Jul/20

$$\frac{1}{(\sqrt{\mathrm{k}}+\sqrt{\mathrm{k}+1})(\sqrt[4]{\mathrm{k}}+\sqrt[4]{\mathrm{k}+1})}\times \frac{\sqrt[4]{\mathrm{k}+1}-\sqrt[4]{\mathrm{k}}}{\sqrt[4]{\mathrm{k}+1}-\sqrt[4]{\mathrm{k}}}=$$$$\frac{\sqrt[4]{\mathrm{k}+1}-\sqrt[4]{\mathrm{k}}}{(\sqrt{\mathrm{k}+1}+\sqrt{\mathrm{k}})(\sqrt{\mathrm{k}+1}-\sqrt{\mathrm{k}})}=\sqrt[4]{\mathrm{k}+1}-\sqrt[4]{\mathrm{k}}$$$$\mathrm{now}\mathrm{S}=\underset{\mathrm{k}=1}{\sum ^{4095}}(\sqrt[4]{\mathrm{k}+1}-\sqrt[4]{\mathrm{k}})=\sqrt[4]{4096}-1=7$$$$\left(\mathrm{note}\mathrm{telecoping}\mathrm{series}\right)$$

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