Let a_ n be a sequence such that a_0=0, a_1=(1)/(2) and 2 a_ n+2 =5 a_ n+1 -3 a_ n , n=0,1,2,3, . Then ∑_ k=1 ^ 100 a_k is equal to

Correct answer: B. 3 a_ 100 -100 aligned & a_0=0, a_1=(1)/(2) & 2 a_ n+2 =5 a_ n+1 -3 a_n & 2 x^2-5 x+3=0 x=1,3 / 2 & a_n=A l^n+B((3)/(2))^n aligned . array ll n=0 & 0=A+B n=1 & (1)/(2)=A+(3)/(2) ~B array ] aligned & A=-1 & ~B=1 aligned aligned & a_ n =-1+((3)/(2))^ n & ∑_ k=1 ^ 100 a_ k =∑_ k=1 ^ 100 (-1)+((3)/(2))^ k aligned aligned & =-100+ ((3)/(2))(((3)/(2))^ 100 -1) (3)/(2)-1 & =-100+3(((3)/(2))^ 100 -1) & =3 ·(a_ 100 )-100 aligned

JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Sequences and Series

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

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

Options

A $3 \mathrm{a}_{99}-100$
B $3 \mathrm{a}_{100}-100$
C $3 \mathrm{a}_{99}+100$
D $3 \mathrm{a}_{100}+100$

Correct answer

B. $3 \mathrm{a}_{100}-100$

Step-by-step solution

\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}

\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}

\begin{aligned} & \Rightarrow \mathrm{a}_{\mathrm{n}}=-1+\left(\frac{3}{2}\right)^{\mathrm{n}} \\ & \sum_{\mathrm{k}=1}^{100} \mathrm{a}_{\mathrm{k}}=\sum_{\mathrm{k}=1}^{100}(-1)+\left(\frac{3}{2}\right)^{\mathrm{k}}\end{aligned}

\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}

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