Linear Recurrence With Little Information

Let us denote sum of $n$ terms is $S(n)$ and $n^\text{th}$ term be denoted by $a(n)$ .

$a(n+3)=[{a(n+2)-a(n+1)}-{a(n+1)-a(n)}]$ . Let us name this equation as $A$ .

Then by taking summation on both sides of equation $A$ we have $S(n+3)=a(n+2)-a(n+1) +a(1)-a(2)$ .

Let this equation be $B$ . Now put $n=1$ in $A$ to get :

$a(4)=a(3)-2a(2)+a(1)$ . Let this equation be $C$ . Now put $n=1$ in $B$ . Then :
$S(4)= a(1)-2a(2)+a(3)$ . Let this equation be $D$ . Now eliminate $a(4)$ from $C$ and $D$ to get $a(2)=-2$ . Now put $n=97$ in $B$ to get $S(100) = 1-a(2)= 3$ .

$\begin{aligned} a_{n+3} & = a_{n+2}-2a_{n+1} + a_n & \small \color{#3D99F6} \text{Given} \\ \sum_{k=1}^n a_{k+3} & = \sum_{k=1}^n \left(a_{k+2}-2a_{k+1} + a_k \right) \\ \sum_{\color{#3D99F6}k=4}^{n+3} a_k & = a_{n+2} - a_{n+1} - a_2 + a_1 & \small \color{#3D99F6} \text{Both sides add }a_1+a_2+a_3 \\ \sum_{\color{#D61F06}k=1}^{n+3} a_k & = a_{n+2} - a_{n+1} + a_3 + 2a_1 & \small \color{#3D99F6} \text{Putting }n=97 \\ \sum_{k=1}^{100} a_k & = a_{99} - a_{98} + a_3 + 2a_1 & \small \color{#3D99F6} \text{Since }a_{98} = a_{99} \\ & = a_3 + 2a_1 & \small \color{#3D99F6} \text{Given that }a_1=a_3 = 1 \\ & = 1+2=\boxed{3} \end{aligned}$

$\text{list}=\text{LinearRecurrence}[\{1,-2,1\},\{1,x,1\},100];$

$\text{list}[[98\text{;;}99]] \Longrightarrow \{437763282635-595866622329 x,1436828120 x-304982319319\} \Longrightarrow x\to \frac{742745601954}{597303450449}$

$\text{Plus}\text{@@}\text{LinearRecurrence}\left[\{1,-2,1\},\left\{1,\frac{742745601954}{597303450449},1\right\},100\right] \longrightarrow 3$