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Question Number 202340 by hardmath last updated on 24/Dec/23

Answered by MATHEMATICSAM last updated on 25/Dec/23

$$\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}$$$$\frac{1}{\sqrt{4}+\sqrt{3}}=\sqrt{4}-\sqrt{3}$$$$\frac{1}{\sqrt{5}+\sqrt{4}}=\sqrt{5}-\sqrt{4}$$$$.$$$$.$$$$.$$$$.$$$$\frac{1}{\sqrt{49}+\sqrt{48}}=\sqrt{49}-\sqrt{48}$$$$$$$$\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{4}}+...+\frac{1}{\sqrt{49}+\sqrt{48}}+\sqrt{2}$$$$=\cancel{\sqrt{3}}-\cancel{\sqrt{2}}+\cancel{\sqrt{4}}-\cancel{\sqrt{3}}+....+\sqrt{49}-\cancel{\sqrt{48}}+\cancel{\sqrt{2}}$$$$=\sqrt{49}=7\left(\mathrm{Ans}\right)$$

Commented by hardmath last updated on 25/Dec/23

$$\mathrm{thankyou}\mathrm{dear}\mathrm{professors}$$

Answered by Frix last updated on 24/Dec/23

$$\frac{1}{\sqrt{k}+\sqrt{k-1}}=\sqrt{k}-\sqrt{k-1}$$$$\underset{k=3}{\sum ^n}(\sqrt{k}-\sqrt{k-1})=\sqrt{n}-\sqrt{2}$$$$\Rightarrow \mathrm{answer}\mathrm{is}\sqrt{49}=7$$

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