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

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Sequences and Series

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, \dots

. Then

\sum_{k = 1}^{100} a_{k}

is equal to :

Answer: (B)

The recurrence

2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}

has characteristic equation

2 x^{2} - 5 x + 3 = 0

, i.e.

(2 x - 3) (x - 1) = 0

, giving roots

x = 1, \frac{3}{2}

.

General solution:

a_{n} = A + B (\frac{3}{2})^{n}

Using

a_{0} = 0

A + B = 0

. Using

a_{1} = \frac{1}{2}

A + \frac{3}{2} B = \frac{1}{2}

.

Solving,

B = 1, A = - 1

, so

a_{n} = - 1 + (\frac{3}{2})^{n}

k = 1 \sum 100 a_{k} = - 100 + k = 1 \sum 100 (\frac{3}{2})^{k} = - 100 + 3 [(\frac{3}{2})^{100} - 1] = 3 (\frac{3}{2})^{100} - 103

Since

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

, we have

(\frac{3}{2})^{100} = a_{100} + 1

, so

k = 1 \sum 100 a_{k} = 3 (a_{100} + 1) - 103 = 3 a_{100} - 100

JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Single correct option