There Are Cancellations?

$\dfrac{1}{a_n}=\dfrac{1}{\sqrt{ 1+\left(1+\dfrac{1}{n} \right)^2 }+\sqrt{ 1+\left(1-\dfrac{1}{n} \right)^2 }}$ Rationalising we get :- $=\dfrac{n\left(\sqrt{ 1+\left(1+\dfrac{1}{n} \right)^2 }-\sqrt{ 1+\left(1-\dfrac{1}{n} \right)^2 }\right)}{4}$ $=\dfrac 14\left(\sqrt{2n^2+2n+1}-\sqrt{2n^2-2n+1}\right)$ We have to find $\mathbf{S}=\displaystyle\sum_{n=1}^{20}\dfrac{1}{a_n}$ $\therefore\large \mathbf{S}=\dfrac 14\left( \begin{aligned} & ~~~~~\left(\cancel{\color{#3D99F6}{\sqrt{5}}}-\sqrt 1\right)\\&+\left(\cancel{\color{#D61F06}{\sqrt{ 13}}}-\cancel{\color{#3D99F6}{\sqrt 5}}\right)\\&+\left(\cancel{\color{#456461}{\sqrt{25}}}-\cancel{\color{#D61F06}{\sqrt{13}}}\right) \\&\cdots\\&\cdots\\&\cdots\\&+\left(\sqrt{841}-\cancel{\color{#EC7300}{\sqrt{761}}}\right)\end{aligned} \right)$ $\color{#302B94}{\mathbf{A ~Telescopic ~Series}}$ $=\dfrac 14\left(\sqrt{841}-\sqrt 1\right)=\Huge\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{\mathbf{7}}}}}}$