Cho $f(n)=\dfrac{4n+\sqrt[]{4n^2-1}}{\sqrt[]{2n+1}+\sqrt[]{2n-1}}$ với $n$ nguyên dương. Tính $f(1)+f(2)+...+f(40).$

Đặt: $(\sqrt{2n-1};\sqrt{2n+1})=(y;x)(y,x\ge 0, x+y \ne 0)$

$\to x^2+y^2=2n-1+2n+1=4n,x^2-y^2=2n+1-2n+1=2\\f(n)=\dfrac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}\\=\dfrac{x^2+xy+y^2}{x+y}\\=\dfrac{(x-y)(x^2+xy+y^2)}{(x-y)(x+y)}\\=\dfrac{x^3-y^3}{x^2-y^2}\\=\dfrac{x^3-y^3}{2}\\=\dfrac{\sqrt{(2n+1)^3}-\sqrt{(2n-1)^3}}{2}$

Ta có:

$f(1)=\dfrac{\sqrt{(2.1+1)^3}-\sqrt{(2.1-1)^3}}{2}=\dfrac{\sqrt{3^3}-\sqrt{1^3}}{2}\\f(2)=\dfrac{\sqrt{5^3}-\sqrt{3^3}}{2}\\...\\f(40)=\dfrac{\sqrt{81^3}-\sqrt{79^3}}{2}\\\to f(1)+...+f(40)=\dfrac{\sqrt{81^3}-\sqrt{1^3}}{2}=\dfrac{728}{2}=364$