Question

Let $f\left(x\right)=\frac{(6x-1)+\sqrt{9x^2-3x-2}}{\sqrt{3x+1}+\sqrt{3x-2}},x\in N$.
If $S=f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(65\right)$, then $\frac{3S}{13}=$

Answer 1

$f\left(x\right)=\frac{\left[\right(3x+1)+(3x-2\left)\right]+\sqrt{9x^2-3x-2}}{\sqrt{3x+1}+\sqrt{3x-2}}$

Put $\sqrt{3x+1}=a$ and $\sqrt{3x-2}=b$

$\Rightarrow f\left(x\right)=\frac{(a^2+b^2+ab)}{a+b}=\frac{a^3-b^3}{a^2-b^2}$
$=\frac{(\sqrt{3x+1}{)}^3-(\sqrt{3x-2}{)}^3}{3}$
Now,
$S=f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(65\right)$
$\Rightarrow S=\frac{(\sqrt{196}{)}^3-1}{3}$
$\Rightarrow 3S=2743$
$\Rightarrow \frac{3S}{13}=211$