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8. Sequences and Series
hard
ધારો કે $\left\langle a_n\right\rangle$ એવી એક શ્રેણી છે કે જેથી $a_0=0, a_1=\frac{1}{2}$ અને $2 a_{n+2}=5 a_{n+1}-3 a_n, n=0,1,2,3, \ldots \ldots$. તો $\sum_{k=1}^{100} a_k$ _______
A$3 a _{99}-100$
B$3 a_{100}-100$
C$3 a _{100}+100$
D$3 a_{99}+100$
(JEE MAIN-2025)
Solution
$a_0=0, a_1=\frac{1}{2}$
$2 a_{n+2}=5 a_{n+1}-3 a_n$
$2 x^2-5 x+3=0 \Rightarrow x=1,3 / 2$
$\therefore a_n=A 1^n+B\left(\frac{3}{2}\right)^n$
$\left.\begin{array}{ll} n =0 & 0= A + B \\ n =1 & \frac{1}{2}= A +\frac{3}{2} B\end{array}\right] \begin{array}{l} A =-1 \\ B=1\end{array}$
$\Rightarrow a_n=-1+\left(\frac{3}{2}\right)^n$
$\sum_{ k =1}^{100} a _{ k }=\sum_{ k =1}^{100}(-1)+\left(\frac{3}{2}\right)^{ k }$
$=-100+\frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100}-1\right)}{\frac{3}{2}-1}$
$=-100+3\left(\left(\frac{3}{2}\right)^{100}-1\right)$
$=3 \cdot\left( a _{100}\right)-100$
$2 a_{n+2}=5 a_{n+1}-3 a_n$
$2 x^2-5 x+3=0 \Rightarrow x=1,3 / 2$
$\therefore a_n=A 1^n+B\left(\frac{3}{2}\right)^n$
$\left.\begin{array}{ll} n =0 & 0= A + B \\ n =1 & \frac{1}{2}= A +\frac{3}{2} B\end{array}\right] \begin{array}{l} A =-1 \\ B=1\end{array}$
$\Rightarrow a_n=-1+\left(\frac{3}{2}\right)^n$
$\sum_{ k =1}^{100} a _{ k }=\sum_{ k =1}^{100}(-1)+\left(\frac{3}{2}\right)^{ k }$
$=-100+\frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100}-1\right)}{\frac{3}{2}-1}$
$=-100+3\left(\left(\frac{3}{2}\right)^{100}-1\right)$
$=3 \cdot\left( a _{100}\right)-100$
Standard 11
Mathematics
