Let a_n be a sequence such that a_0=0,a_1=(1)/(2) and 2a_n+2=5a_n+1-3a_n,n=0,1,2,3,.

Question

Let $\langle {a}_{n}\rangle $ be a sequence such that ${a}_{0}=0,{a}_{1}=\frac{1}{2}$ and $2{a}_{n+2}=5{a}_{n+1}-3{a}_{n},n=0,1,2,3,\ldots$. Then $\sum _{k=1}^{100}{a}_{k}$ is equal to

Accepted Answer

$$\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}\end{aligned}$$\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
\]

$$\begin{aligned}
& =-100+\frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100}-1\right)}{\frac{3}{2}-1} \
& =-100+3\left(\left(\frac{3}{2}\right)^{100}-1\right) \
& =3 \cdot\left(\mathrm{a}_{100}\right)-100
\end{aligned}$$

Solutions · 1