Question

please solbe these math question

Answer 1

Answer:

$\displaystyle\frac{\sqrt{2n+3}-\sqrt3}{2}$

Step-by-step explanation:

The k-th term of this series is

$\displaystyle\frac1{\sqrt{2k+1}+\sqrt{2k+3}}\\\\\\=\frac{\sqrt{2k+3}-\sqrt{2k+1}}{\bigl(\sqrt{2k+3}+\sqrt{2k+1}\bigr)\bigl(\sqrt{2k+3}-\sqrt{2k+1}\bigr)}\\\\\\=\frac{\sqrt{2k+3}-\sqrt{2k+1}}{(2k+3)-(2k+1)}\\\\\\=\frac{\sqrt{2k+3}-\sqrt{2k+1}}{2}\\\\=\tfrac12\sqrt{2k+3} - \tfrac12\sqrt{2k+1}$

Therefore, the sum of the first n terms is:

$\displaystyle\Bigl(\tfrac12\sqrt{5} - \tfrac12\sqrt{3}\Bigr)+\Bigl(\tfrac12\sqrt{7} - \tfrac12\sqrt{5}\Bigr)+\dots+\Bigl(\tfrac12\sqrt{2n+3} - \tfrac12\sqrt{2n+1}\Bigr)\\\\=-\tfrac12\sqrt3 + \tfrac12\sqrt{2n+3}\\\\=\frac{\sqrt{2n+3}-\sqrt3}{2}$