Let <an> be a sequence such that a0=0, a1=1/2 and 2an+2=5an+1-3an,

Question

Let \(<a_{n}>\) 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 \(\sum\limits_{k = 1}^{100}a_k\) is equal to :

(1) \(3 a_{99}-100 \)

(2) \(3 a_{100}-100\)

(3) \(3 a_{100}+100\)

(4) \(3 \mathrm{a}_{99}+100\)

Answer 1

Correct option is (2) \(3 a_{100}-100\)  

\(a_{0}=0, a_{1}=\frac{1}{2}\)

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

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

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

\(\left.\begin{array}{ll}\mathrm{n}=0 & 0=\mathrm{A}+\mathrm{B} \\ \mathrm{n}=1 & \frac{1}{2}=\mathrm{A}+\frac{3}{2} \mathrm{~B}\end{array}\right] \begin{aligned} & \mathrm{A}=-1 \\ & \mathrm{~B}=1\end{aligned}\)

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

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

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

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

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