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Substituting $n=k-1$ into the given equation yields

$\sqrt{a_1}+\sqrt{a_2 -1}+\sqrt{a_3-2}+...+\sqrt{a_{k-1}-(k-2)} = \dfrac{a_1+a_2+a_3+...+a_{k-1}}{2}-\dfrac{(k-1)(k-4)}{4}$

Substituting $n=k$ into the given equation yields

$\sqrt{a_1}+\sqrt{a_2 -1}+\sqrt{a_3-2}+...+\sqrt{a_{k}-(k-1)} = \dfrac{a_1+a_2+a_3+...+a_{k}}{2}-\dfrac{k(k-3)}{4}$

Subtracting the first equation from the second equation yields

$\sqrt{a_{k}-(k-1)} = \dfrac{a_{k}}{2}-\dfrac{k(k-3)-(k-1)(k-4)}{4}$

$2\sqrt{a_{k}-(k-1)} = a_{k} -(k-2)$

$4[a_{k}-(k-1)] = a^{2}_{k} - 2(a_{k})(k-2) + (k^{2}-4k+4)$

$a^{2}_{k} - 2(a_{k})(k) + k^{2} = 0$

$(a_{k} - k)^{2} = 0$

So $a_{k} = k$ .

Therefore, $a_{50}-a_{51}+a_{52}-a_{53}+...+a_{100} = (50-51)+(52-53)+(54-55)+...+(98-99)+100 = 25(-1)+100 = 75$ .

Check the value of $a_1$ by putting $n=1$

$a_1=1$

Using this and $n=2$ , check the value of $a_2$ . It comes out to be 2.

Same way, if you check all the other values,

$a_3=3$

$a_4=4$ ...and so on.

Hence, $a_n=n$ , making the required sum $100+25(-1)=75$

(This solution was by complete experimentation, i.e. just finding out values of a few $a_i$ and making a conclusion, I have no idea how to do this theoretically)