Let langle a_ n rangle be a sequence such that a_0=0, a_1= 1 2 and 2 a_ n+2=5 a_ n+1-3 a_ n, n=0,1,2,3, …. Then Σ_ k=1^100 a_k is equal to

$\begin{aligned} & a_0=0, a_1=\frac{1}{2} \\ & 2 a_{n+2}=5 a_{n+1}-3 a_n \\ & 2 x^2-5 x+3=0 \Rightarrow x=1,3 / 2 \\ & \therefore a_n=A l^n+B\left(\frac{3}{2}\right)^n\end{aligned}$​a0​=0,a1​=21​2an+2​=5an+1​−3an​2x2−5x+3=0⇒x=1,3/2∴an​=Aln+B(23​)n​ $\left.\begin{array}{ll}\mathrm{n}=0 & 0=\mathrm{A}+\mathrm{B} \\ \mathrm{n}=1 & \frac{1}{2}=\mathrm{A}+\frac{3}{2} \mathrm{~B}\end{array}\right] \begin{aligned} & \mathrm{A}=-1 \\ & \mathrm{~B}=1\end{aligned}$n=0n=1​0=A+B21​=A+23​ B​]​A=−1 B=1​ $\displaystyle \begin{aligned} & \Rightarrow \mathrm{a}_{\mathrm{n}}=-1+\left(\frac{3}{2}\right)^{\mathrm{n}} \\ & \sum_{\mathrm{k}=1}^{100} \mathrm{a}_{\mathrm{k}}=\sum_{\mathrm{k}=1}^{100}(-1)+\left(\frac{3}{2}\right)^{\mathrm{k}}\end{aligned}$​⇒an​=−1+(23​)nk=1∑100​ak​=k=1∑100​(−1)+(23​)k​ $\begin{aligned} & =-100+\frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100}-1\right)}{\frac{3}{2}-1} \\ & =-100+3\left(\left(\frac{3}{2}\right)^{100}-1\right) \\ & =3 \cdot\left(\mathrm{a}_{100}\right)-100\end{aligned}$​=−100+23​−1(23​)((23​)100−1)​=−100+3((23​)100−1)=3⋅(a100​)−100​