Let $\left\langle a_{\mathrm{n}}\right\rangle$ be a sequence such that $a_0=0, a_1=\frac{1}{2}$ and $2 a_{\mathrm{n}+2}=5 a_{\mathrm{n}+1}-3 a_{\mathrm{n}}, \mathrm{n}=0,1,2,3, \ldots$. Then $\sum\limits_{k=1}^{100} a_k$ is equal to

<p>$$\begin{aligned} & a_0=0, a_1=\frac{1}{2} \\ & 2 a_{n+2}=5 a_{n+1}-3 a_n \\ & 2 x^2-5 x+3=0 \Rightarrow x=1,3 / 2 \\ & \therefore a_n=A(1)^n+B\left(\frac{3}{2}\right)^n \\ &\left.\begin{array}{cc} \mathrm{n}=0 & 0=\mathrm{A}+\mathrm{B} \\ \mathrm{n}=1 & \frac{1}{2}=\mathrm{A}+\frac{3}{2} \mathrm{~B} \end{array}\right] \begin{aligned} & \mathrm{A}=-1 \\ & \mathrm{~B}=1 \end{aligned}\\ & \Rightarrow a_n=-1+\left(\frac{3}{2}\right)^n \\ & \sum_{k=1}^{100} a_k=\sum_{k=1}^{100}(-1)+\left(\frac{3}{2}\right)^k \end{aligned}$$</p> <p>$$\begin{aligned} & =-100+\frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100}-1\right)}{\frac{3}{2}-1} \\ & =-100+3\left(\left(\frac{3}{2}\right)^{100}-1\right) \\ & =3 \cdot\left(\mathrm{a}_{100}\right)-100 \end{aligned}$$</p>