jmc

Writing down the recursion for $n=1,2,3,\dots ,97,$ we have $$\begin{array}{rl}{a}_4& =a_3-2a_2+a_1\\ a_5& =a_4-2a_3+a_2\\ a_6& =a_5-2a_4+a_3\\ & ⋮\\ a_{100}& =a_{99}-2a_{98}+a_{97}.\end{array}$$Summing all $97$ of these equations, we have $$a_4+\cdots +a_{100}=(a_3+\cdots +a_{99})-2(a_2+\cdots +a_{98})+(a_1+\cdots +a_{97}).$$Let $S=a_1+a_2+\cdots +a_{100}.$ Then we can rewrite the above equation in terms of $S$ as $$S-(a_1+a_2+a_3)=[S-(a_1+a_2+a_{100})]-2[S-(a_1+a_{99}+a_{100})]+[S-(a_{98}+a_{99}+a_{100})],$$or $$S-a_1-a_2-a_3=a_1-a_2-a_{98}+a_{99}.$$Thus, $$S=2a_1+a_3-a_{98}+a_{99}.$$Since $a_1=a_3=1$ and $a_{98}=a_{99},$ we get $$S=2(1)+1=\boxed{3}.$$