Solution

Let $$\sqrt{a_1-1} = b_1$$;
$$\sqrt{a_2 - 1} = b_2$$
$$\sqrt{a_3 - 2} = b_3$$
$$\dots \dots$$
$$\sqrt{a_n - (n-1)} = b_n$$

$$\therefore $$ $$ b_1 + b_2 + \dots + b_n =\frac{1}{2} \Big[ b_1^2 + (b_2^2 + 1) + (b_3^2 + 2) + \dots + (b_n^2 + (n-1)) \Big] - \dfrac{n(n-3)}{4}$$

$$\therefore $$ $$ \displaystyle \sum b_i = \frac{1}{2}[ (b_1^2 + b_2^2 + b_3^2 + \dots + b_n^2) + ( 1+ 2+3 + \dots + (n-1)] - \dfrac{n(n-3)}{4}$$

$$\displaystyle \Rightarrow 2 \sum b_i = \sum b_i^2 + \frac{n(n-1)}{2} - \frac{n(n-3)}{2} $$

$$\displaystyle \Rightarrow 2 \sum b_i = \sum b_i^2 + n$$

$$\therefore $$ $$ \displaystyle \sum b_i^2 - 2 \sum b_i + \sum 1 = 0$$

$$\displaystyle \Rightarrow 2 \sum_{i = 1}^{n} (b_i - 1)^2 =0$$

$$b_1 - 1 = 0 \Rightarrow b_1^2 = a_1 = 1$$

$$b_2 - 1 = 0 \Rightarrow b_2^2 = a_2 - 1 = 1 \Rightarrow a_2 = 2$$

$$b_3- 1 = 0 \Rightarrow b_3^2 = a_3 -2 = 1 \Rightarrow a_3 = 3$$ and so on

Hence $$a_n = n$$

$$\displaystyle \therefore \ \ \ 2 \sum_{i = 1}^{100} a_i =1 + 2 + 3 + \dots + 100 = 5050$$