What is the value of $$\frac{5}{\sqrt{2}+1}+\frac{5}{\sqrt{3}+\sqrt{2}}+\frac{5}{\sqrt{4}+\sqrt{3}}+....\frac{5}{\sqrt{121}+\sqrt{120}}$$ ?

Expression : $$\frac{5}{\sqrt{2}+1}+\frac{5}{\sqrt{3}+\sqrt{2}}+\frac{5}{\sqrt{4}+\sqrt{3}}+....\frac{5}{\sqrt{121}+\sqrt{120}}$$

Rationalizing the denominator, we get :

= $$(\frac{5}{\sqrt{2}+1}\times\frac{\sqrt2-1}{\sqrt2-1})+(\frac{5}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt3-2}{\sqrt3-2})+(\frac{5}{\sqrt{4}+\sqrt{3}}\times\frac{\sqrt4-\sqrt3}{\sqrt4-\sqrt3})+....+(\frac{5}{\sqrt{121}+\sqrt{120}}\times\frac{\sqrt{121}-\sqrt{120}}{\sqrt{121}-\sqrt{120}})$$

Using, $$(a-b)(a+b)=a^2-b^2$$

= $$\frac{5\times(\sqrt2-1)}{(2-1)}+\frac{5\times(\sqrt3-\sqrt2)}{(3-2)}+\frac{5\times(\sqrt4-\sqrt3)}{(4-3)}+....+\frac{5\times(\sqrt{121}-\sqrt{120})}{(121-120)}$$

Thus, after cancelling the positive and negative terms alternatively, we are left with :

= $$-5+5(\sqrt{121})=-5+55=50$$

=> Ans - (B)