Đáp án: `S=\frac{(n+1)\sqrt{n+1}-1}{2}`
Giải thích các bước giải:
Xét biểu thức tổng quát:
`\frac{2n+\sqrt{n^2-1}}{\sqrt{n-1}+\sqrt{n+1}}=\frac{(\sqrt{n+1}-\sqrt{n-1})[(n+1)+\sqrt{n+1}.\sqrt{n-1}+(n-1)]}{(n+1)-(n-1)}`
`=\frac{(\sqrt{n+1})^3-(\sqrt{n-1})^3}{2}`
Như vậy:
`\frac{4+\sqrt{3}}{\sqrt{1}+\sqrt{3}}=\frac{(\sqrt{3})^3-(\sqrt{1})^3}{2}`
`\frac{8+\sqrt{15}}{\sqrt{3}+\sqrt{5}}=\frac{(\sqrt{5})^3-(\sqrt{3})^3}{2}`
`\frac{12+\sqrt{35}}{\sqrt{5}+\sqrt{7}}=\frac{(\sqrt{7})^3-(\sqrt{5})^3}{2}`
`⇒S=\frac{(\sqrt{3})^3-(\sqrt{1})^3}{2}+\frac{(\sqrt{5})^3-(\sqrt{3})^3}{2}+\frac{(\sqrt{7})^3-(\sqrt{5})^3}{2}+.....+\frac{(\sqrt{n+1})^3-(\sqrt{n-1})^3}{2}`
`=\frac{(\sqrt{n+1})^3-(\sqrt{1})^3}{2}=\frac{(n+1)\sqrt{n+1}-1}{2}`
