Question

1/1+root2+1/rrot2+root3+1/root3+root4+1/root4+root5+1/root5+root6+1/root6+root7+1/root7 + root8+1/root8+root9 = 2 . prove that
please help me

Answer 1

Notice that
$\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\times\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}^2-\sqrt{n}^2}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}}$

The given sum is same as
$(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\cdots+(\sqrt{8}-\sqrt{7})+(\sqrt{9}-\sqrt{8})\\\text{cancel the~middle terms and get}\\=-\sqrt{1}+\sqrt{9}\\=-1+3\\=2$

Answer 2

Hope the attachment helps you