JEE Main 2025 — 28 January, Morning Shift — Question 9

$a_{0}=0, a_{1}=\frac{1}{2}$

$2 \mathrm{a}_{\mathrm{n}+2}=5 \mathrm{a}_{\mathrm{n}+1}-3 \mathrm{a}_{\mathrm{n}}$

$2 \mathrm{x}^{2}-5 \mathrm{x}+3=0 \Rightarrow \mathrm{x}=1,3 / 2$

$\therefore \mathrm{a}_{\mathrm{n}}=\mathrm{A} (1)^{\mathrm{n}}+\mathrm{B}\left(\frac{3}{2}\right)^{\mathrm{n}}$

$\left.\begin{array}{cc}\mathrm{n}=0, & 0=\mathrm{A}+\mathrm{B} ;\mathrm{n}=1, & \frac{1}{2}=\mathrm{A}+\frac{3}{2} \mathrm{~B}\end{array}\right] \begin{gathered}\mathrm{A}=-1 ,\mathrm{~B}=1\end{gathered}$

$\Rightarrow \mathrm{a}_{\mathrm{n}}=-1+\left(\frac{3}{2}\right)^{\mathrm{n}}$

$\sum_{k=1}^{100} a_{k}=\sum_{k=1}^{100}(-1)+\left(\frac{3}{2}\right)^{k}$

$=-100+\frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100}-1\right)}{\frac{3}{2}-1}$

$=-100+3\left(\left(\frac{3}{2}\right)^{100}-1\right)$

$=3 .\left(\mathrm{a}_{100}\right)-100$