Question

The value of ${\sum }_{r=1}^n\frac{1}{\sqrt{a+rx}+\sqrt{a+(r-1)x}}$ is

• A. $\frac{n}{\sqrt{a}+\sqrt{a+nx}}$
• B. $\frac{\sqrt{a+nx}-\sqrt{a}}{x}$
• C. $\frac{n(\sqrt{a+nx}-a)}{x}$
• D. none of these

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{n}{\sqrt{a}+\sqrt{a+nx}}$
B $\frac{\sqrt{a+nx}-\sqrt{a}}{x}$

Let $S={\sum }_{r=1}^n\frac{1}{\sqrt{a+rx}+\sqrt{a+(r-1)x}}$
On rationalizing denominator ,we get
$S=-\frac{1}{x}{\sum }_{r=1}^n(\sqrt{a+(r-1)x}-\sqrt{a+rx})$
$\Rightarrow S=-\frac{1}{x}[\sqrt{a}-\sqrt{a+x}+\sqrt{a+x}-\sqrt{a+2x}+......-\sqrt{a+nx}]$
$\Rightarrow S=\frac{\sqrt{a+nx}-\sqrt{a}}{x}$
On rationalizing numerator, we get
$S=\frac{n}{\sqrt{a}+\sqrt{a+nx}}$
Ans: A,B