Mathematics - Sequence & Series Question with Solution | TestHub

$\begin{array}{rl}& a_0=0,a_1=\frac{1}{2}\\ & 2a_{n+2}=5a_{n+1}-3a_n\\ & 2x^2-5x+3=0\Rightarrow x=1,3/2\\ & \therefore a_n=A{l}^n+B{\left(\frac{3}{2}\right)}^n\end{array}$ $\begin{array}{ll}\mathrm{n}=0& 0=\mathrm{A}+\mathrm{B}\\ \mathrm{n}=1& \frac{1}{2}=\mathrm{A}+\frac{3}{2}\mathrm{B}\end{array}]\begin{array}{rl}& \mathrm{A}=-1\\ & \mathrm{B}=1\end{array}$ $\begin{array}{rl}& \Rightarrow {\mathrm{a}}_{\mathrm{n}}=-1+{\left(\frac{3}{2}\right)}^{\mathrm{n}}\\ & \sum _{\mathrm{k}=1}^{100}{\mathrm{a}}_{\mathrm{k}}=\sum _{\mathrm{k}=1}^{100}(-1)+{\left(\frac{3}{2}\right)}^{\mathrm{k}}\end{array}$ $\begin{array}{rl}& =-100+\frac{\left(\frac{3}{2}\right)\left({\left(\frac{3}{2}\right)}^{100}-1\right)}{\frac{3}{2}-1}\\ & =-100+3\left({\left(\frac{3}{2}\right)}^{100}-1\right)\\ & =3\cdot \left({\mathrm{a}}_{100}\right)-100\end{array}$