Let leftlangle a_nrightrangle be a sequence s | vedclass

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Question

$a_0=0, a_1=\frac{1}{2}$
$2 a_{n+2}=5 a_{n+1}-3 a_n$
$2 x^2-5 x+3=0 \Rightarrow x=1,3 / 2$
$	herefore a_n=A 1^n+B\left(\frac{3}{2}ight)^n$
$\le

Vedclass · Accepted Answer

$3 a_{100}-100$

Vedclass · Answer

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

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

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