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Question

The sum to $$n$$ terms of series $$\displaystyle\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+....$$.


Solution

On inspection and further simplification, the general term of the series is found to be:
$$T_n = \dfrac{1}{\sqrt{2n-1}+\sqrt{2n+1}} \\= \dfrac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\times\dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{\sqrt{2n+1}-\sqrt{2n-1}} \\=  \dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{(2n+1)-(2n-1)} = \dfrac{\sqrt{2n+1}-\sqrt{2n-1}}{2}$$
The sum to n terms is then given by:
$$\sum T_n = T_n + T_{n-1} + .... + T_1 = \dfrac{\sqrt{2n+1}-\sqrt{1}}{2} = \dfrac{\sqrt{2n+1}-1}{2}$$
Note that due to the nature of the general term, the intermediate portions of the sum cancel all the way to the very first term, whose negative part is retained in the final answer.

Mathematics

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