JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(\left\langle a_{\mathrm{n}}\right\rangle\) be a sequence such that \(a_0=0, a_1=\frac{1}{2}\) and \(2 a_{\mathrm{n}+2}=5 a_{\mathrm{n}+1}-3 a_{\mathrm{n}}, \mathrm{n}=0,1,2,3, \ldots\). Then \(\sum_{\mathrm{k}=1}^{100} a_k\) is equal to
- A \(3 \mathrm{a}_{99}-100\)
- B \(3 \mathrm{a}_{100}-100\)
- C \(3 \mathrm{a}_{99}+100\)
- D \(3 \mathrm{a}_{100}+100\)
Answer & Solution
Correct Answer
(B) \(3 \mathrm{a}_{100}-100\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & a_0=0, a_1=\frac{1}{2} \\ & 2 a_{n+2}=5 a_{n+1}-3 a_n \\ & 2 x^2-5 x+3=0 \Rightarrow x=1,3 / 2 \\ & \therefore a_n=A l^n+B\left(\frac{3}{2}\right)^n\end{aligned}\)…
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